partial fractions homework

7.4 Partial Fractions

Learning objectives.

In this section, you will:

• Decompose   P ( x ) Q ( x ) P ( x ) Q ( x ) , where   Q ( x ) Q ( x ) has only nonrepeated linear factors.
• Decompose   P ( x ) Q ( x ) P ( x ) Q ( x ) , where   Q ( x ) Q ( x )   has repeated linear factors.
• Decompose   P ( x ) Q ( x ) P ( x ) Q ( x ) , where   Q ( x ) Q ( x ) has a nonrepeated irreducible quadratic factor.
• Decompose   P ( x ) Q ( x ) P ( x ) Q ( x ) , where   Q ( x ) Q ( x ) has a repeated irreducible quadratic factor.

Earlier in this chapter, we studied systems of two equations in two variables, systems of three equations in three variables, and nonlinear systems. Here we introduce another way that systems of equations can be utilized—the decomposition of rational expressions.

Fractions can be complicated; adding a variable in the denominator makes them even more so. The methods studied in this section will help simplify the concept of a rational expression.

Decomposing P ( x ) Q ( x ) P ( x ) Q ( x ) Where Q(x) Has Only Nonrepeated Linear Factors

Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at partial fraction decomposition , which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified rational expression to the original expressions, called the partial fraction .

For example, suppose we add the following fractions:

We would first need to find a common denominator, ( x + 2 ) ( x −3 ) . ( x + 2 ) ( x −3 ) .

Next, we would write each expression with this common denominator and find the sum of the terms.

Partial fraction decomposition is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions.

We will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator.

When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of x 2 − x −6 x 2 − x −6 are ( x −3 ) ( x + 2 ) , ( x −3 ) ( x + 2 ) , the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition.

Partial Fraction Decomposition of P ( x ) Q ( x ) : Q ( x ) P ( x ) Q ( x ) : Q ( x ) Has Nonrepeated Linear Factors

The partial fraction decomposition of P ( x ) Q ( x ) P ( x ) Q ( x ) when Q ( x ) Q ( x ) has nonrepeated linear factors and the degree of P ( x ) P ( x ) is less than the degree of Q ( x ) Q ( x ) is

Given a rational expression with distinct linear factors in the denominator, decompose it.

• Use a variable for the original numerators, usually A , B ,  A , B ,  or C , C , depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use A n A n for each numerator P ( x ) Q ( x ) = A 1 ( a 1 x + b 1 ) + A 2 ( a 2 x + b 2 ) + ⋯ + A n ( a n x + b n ) P ( x ) Q ( x ) = A 1 ( a 1 x + b 1 ) + A 2 ( a 2 x + b 2 ) + ⋯ + A n ( a n x + b n )
• Multiply both sides of the equation by the common denominator to eliminate fractions.
• Expand the right side of the equation and collect like terms.
• Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.

Decomposing a Rational Function with Distinct Linear Factors

Decompose the given rational expression with distinct linear factors.

We will separate the denominator factors and give each numerator a symbolic label, like A , B , A , B , or C . C .

Multiply both sides of the equation by the common denominator to eliminate the fractions:

The resulting equation is

Set up a system of equations associating corresponding coefficients.

Add the two equations and solve for B . B .

Substitute B = 1 B = 1 into one of the original equations in the system.

Thus, the partial fraction decomposition is

Another method to use to solve for A A or B B is by considering the equation that resulted from eliminating the fractions and substituting a value for x x that will make either the A - or B -term equal 0. If we let x = 1 , x = 1 , the A - A - term becomes 0 and we can simply solve for B . B .

Next, either substitute B = 1 B = 1 into the equation and solve for A , A , or make the B -term 0 by substituting x = −2 x = −2 into the equation.

We obtain the same values for A A and B B using either method, so the decompositions are the same using either method.

Although this method is not seen very often in textbooks, we present it here as an alternative that may make some partial fraction decompositions easier. It is known as the Heaviside method , named after Charles Heaviside, a pioneer in the study of electronics.

Find the partial fraction decomposition of the following expression.

Decomposing P ( x ) Q ( x ) P ( x ) Q ( x ) Where Q(x) Has Repeated Linear Factors

Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.

Partial Fraction Decomposition of P ( x ) Q ( x ) : Q ( x ) P ( x ) Q ( x ) : Q ( x ) Has Repeated Linear Factors

The partial fraction decomposition of P ( x ) Q ( x ) , P ( x ) Q ( x ) , when Q ( x ) Q ( x ) has a repeated linear factor occurring n n times and the degree of P ( x ) P ( x ) is less than the degree of Q ( x ) , Q ( x ) , is

Write the denominator powers in increasing order.

Given a rational expression with repeated linear factors, decompose it.

• Use a variable like A , B , A , B , or C C for the numerators and account for increasing powers of the denominators. P ( x ) Q ( x ) = A 1 ( a x + b ) + A 2 ( a x + b ) 2 +   .  .  . +  A n ( a x + b ) n P ( x ) Q ( x ) = A 1 ( a x + b ) + A 2 ( a x + b ) 2 +   .  .  . +  A n ( a x + b ) n

Decomposing with Repeated Linear Factors

Decompose the given rational expression with repeated linear factors.

The denominator factors are x ( x −2 ) 2 . x ( x −2 ) 2 . To allow for the repeated factor of ( x −2 ) , ( x −2 ) , the decomposition will include three denominators: x , ( x −2 ) , x , ( x −2 ) , and ( x −2 ) 2 . ( x −2 ) 2 . Thus,

Next, we multiply both sides by the common denominator.

On the right side of the equation, we expand and collect like terms.

Next, we compare the coefficients of both sides. This will give the system of equations in three variables:

Solving for A A , we have

Substitute A = 1 A = 1 into equation (1).

Then, to solve for C , C , substitute the values for A A and B B into equation (2).

Find the partial fraction decomposition of the expression with repeated linear factors.

Decomposing P ( x ) Q ( x ) , P ( x ) Q ( x ) , Where Q(x) Has a Nonrepeated Irreducible Quadratic Factor

So far, we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators A , B , A , B , or C C representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as A x + B , B x + C , A x + B , B x + C , etc.

Decomposition of P ( x ) Q ( x ) : Q ( x ) P ( x ) Q ( x ) : Q ( x ) Has a Nonrepeated Irreducible Quadratic Factor

The partial fraction decomposition of P ( x ) Q ( x ) P ( x ) Q ( x ) such that Q ( x ) Q ( x ) has a nonrepeated irreducible quadratic factor and the degree of P ( x ) P ( x ) is less than the degree of Q ( x ) Q ( x ) is written as

The decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator: A , B , C , A , B , C , and so on.

Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it.

• Use variables such as A , B , A , B , or C C for the constant numerators over linear factors, and linear expressions such as A 1 x + B 1 , A 2 x + B 2 , A 1 x + B 1 , A 2 x + B 2 , etc., for the numerators of each quadratic factor in the denominator. P ( x ) Q ( x ) = A a x + b + A 1 x + B 1 ( a 1 x 2 + b 1 x + c 1 ) + A 2 x + B 2 ( a 2 x 2 + b 2 x + c 2 ) + ⋅ ⋅ ⋅ + A n x + B n ( a n x 2 + b n x + c n ) P ( x ) Q ( x ) = A a x + b + A 1 x + B 1 ( a 1 x 2 + b 1 x + c 1 ) + A 2 x + B 2 ( a 2 x 2 + b 2 x + c 2 ) + ⋅ ⋅ ⋅ + A n x + B n ( a n x 2 + b n x + c n )

Decomposing P ( x ) Q ( x ) P ( x ) Q ( x ) When Q(x) Contains a Nonrepeated Irreducible Quadratic Factor

Find a partial fraction decomposition of the given expression.

We have one linear factor and one irreducible quadratic factor in the denominator, so one numerator will be a constant and the other numerator will be a linear expression. Thus,

We follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator.

Notice we could easily solve for A A by choosing a value for x x that will make the B x + C B x + C term equal 0. Let x = −3 x = −3 and substitute it into the equation.

Now that we know the value of A , A , substitute it back into the equation. Then expand the right side and collect like terms.

Setting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations.

Solve for B B using equation (1) and solve for C C using equation (3).

Thus, the partial fraction decomposition of the expression is

Could we have just set up a system of equations to solve Example 3 ?

Yes, we could have solved it by setting up a system of equations without solving for A A first. The expansion on the right would be:

So the system of equations would be:

Find the partial fraction decomposition of the expression with a nonrepeating irreducible quadratic factor.

Decomposing P ( x ) Q ( x ) P ( x ) Q ( x ) When Q(x) Has a Repeated Irreducible Quadratic Factor

Now that we can decompose a simplified rational expression with an irreducible quadratic factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers.

Decomposition of P ( x ) Q ( x ) P ( x ) Q ( x ) When Q(x) Has a Repeated Irreducible Quadratic Factor

The partial fraction decomposition of P ( x ) Q ( x ) , P ( x ) Q ( x ) , when Q ( x ) Q ( x ) has a repeated irreducible quadratic factor and the degree of P ( x ) P ( x ) is less than the degree of Q ( x ) , Q ( x ) , is

Write the denominators in increasing powers.

Given a rational expression that has a repeated irreducible factor, decompose it.

• Use variables like A , B , A , B , or C C for the constant numerators over linear factors, and linear expressions such as A 1 x + B 1 , A 2 x + B 2 , A 1 x + B 1 , A 2 x + B 2 , etc., for the numerators of each quadratic factor in the denominator written in increasing powers, such as P ( x ) Q ( x ) = A a x + b + A 1 x + B 1 ( a x 2 + b x + c ) + A 2 x + B 2 ( a x 2 + b x + c ) 2 + ⋯ + A n + B n ( a x 2 + b x + c ) n P ( x ) Q ( x ) = A a x + b + A 1 x + B 1 ( a x 2 + b x + c ) + A 2 x + B 2 ( a x 2 + b x + c ) 2 + ⋯ + A n + B n ( a x 2 + b x + c ) n

Decomposing a Rational Function with a Repeated Irreducible Quadratic Factor in the Denominator

Decompose the given expression that has a repeated irreducible factor in the denominator.

The factors of the denominator are x , ( x 2 + 1 ) , x , ( x 2 + 1 ) , and ( x 2 + 1 ) 2 . ( x 2 + 1 ) 2 . Recall that, when a factor in the denominator is a quadratic that includes at least two terms, the numerator must be of the linear form A x + B . A x + B . So, let’s begin the decomposition.

We eliminate the denominators by multiplying each term by x ( x 2 + 1 ) 2 . x ( x 2 + 1 ) 2 . Thus,

Expand the right side.

Now we will collect like terms.

Set up the system of equations matching corresponding coefficients on each side of the equal sign.

We can use substitution from this point. Substitute A = 1 A = 1 into the first equation.

Substitute A = 1 A = 1 and B = 0 B = 0 into the third equation.

Substitute C = 1 C = 1 into the fourth equation.

Now we have solved for all of the unknowns on the right side of the equal sign. We have A = 1 , A = 1 , B = 0 , B = 0 , C = 1 , C = 1 , D = −1 , D = −1 , and E = −2. E = −2. We can write the decomposition as follows:

Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor.

Access these online resources for additional instruction and practice with partial fractions.

• Partial Fraction Decomposition
• Partial Fraction Decomposition With Repeated Linear Factors
• Partial Fraction Decomposition With Linear and Quadratic Factors

7.4 Section Exercises

Can any quotient of polynomials be decomposed into at least two partial fractions? If so, explain why, and if not, give an example of such a fraction

Can you explain why a partial fraction decomposition is unique? (Hint: Think about it as a system of equations.)

Can you explain how to verify a partial fraction decomposition graphically?

You are unsure if you correctly decomposed the partial fraction correctly. Explain how you could double-check your answer.

Once you have a system of equations generated by the partial fraction decomposition, can you explain another method to solve it? For example if you had 7 x + 13 3 x 2 + 8 x + 15 = A x + 1 + B 3 x + 5 7 x + 13 3 x 2 + 8 x + 15 = A x + 1 + B 3 x + 5 , we eventually simplify to 7 x + 13 = A ( 3 x + 5 ) + B ( x + 1 ) . 7 x + 13 = A ( 3 x + 5 ) + B ( x + 1 ) . Explain how you could intelligently choose an x x -value that will eliminate either A A or B B and solve for A A and B . B .

For the following exercises, find the decomposition of the partial fraction for the nonrepeating linear factors.

5 x + 16 x 2 + 10 x + 24 5 x + 16 x 2 + 10 x + 24

3 x −79 x 2 −5 x −24 3 x −79 x 2 −5 x −24

− x −24 x 2 −2 x −24 − x −24 x 2 −2 x −24

10 x + 47 x 2 + 7 x + 10 10 x + 47 x 2 + 7 x + 10

x 6 x 2 + 25 x + 25 x 6 x 2 + 25 x + 25

32 x −11 20 x 2 −13 x + 2 32 x −11 20 x 2 −13 x + 2

x + 1 x 2 + 7 x + 10 x + 1 x 2 + 7 x + 10

5 x x 2 −9 5 x x 2 −9

10 x x 2 −25 10 x x 2 −25

6 x x 2 −4 6 x x 2 −4

2 x −3 x 2 −6 x + 5 2 x −3 x 2 −6 x + 5

4 x −1 x 2 − x −6 4 x −1 x 2 − x −6

4 x + 3 x 2 + 8 x + 15 4 x + 3 x 2 + 8 x + 15

3 x −1 x 2 −5 x + 6 3 x −1 x 2 −5 x + 6

For the following exercises, find the decomposition of the partial fraction for the repeating linear factors.

−5 x −19 ( x + 4 ) 2 −5 x −19 ( x + 4 ) 2

x ( x −2 ) 2 x ( x −2 ) 2

7 x + 14 ( x + 3 ) 2 7 x + 14 ( x + 3 ) 2

−24 x −27 ( 4 x + 5 ) 2 −24 x −27 ( 4 x + 5 ) 2

−24 x −27 ( 6 x −7 ) 2 −24 x −27 ( 6 x −7 ) 2

5 − x ( x −7 ) 2 5 − x ( x −7 ) 2

5 x + 14 2 x 2 + 12 x + 18 5 x + 14 2 x 2 + 12 x + 18

5 x 2 + 20 x + 8 2 x ( x + 1 ) 2 5 x 2 + 20 x + 8 2 x ( x + 1 ) 2

4 x 2 + 55 x + 25 5 x ( 3 x + 5 ) 2 4 x 2 + 55 x + 25 5 x ( 3 x + 5 ) 2

54 x 3 + 127 x 2 + 80 x + 16 2 x 2 ( 3 x + 2 ) 2 54 x 3 + 127 x 2 + 80 x + 16 2 x 2 ( 3 x + 2 ) 2

x 3 −5 x 2 + 12 x + 144 x 2 ( x 2 + 12 x + 36 ) x 3 −5 x 2 + 12 x + 144 x 2 ( x 2 + 12 x + 36 )

For the following exercises, find the decomposition of the partial fraction for the irreducible nonrepeating quadratic factor.

4 x 2 + 6 x + 11 ( x + 2 ) ( x 2 + x + 3 ) 4 x 2 + 6 x + 11 ( x + 2 ) ( x 2 + x + 3 )

4 x 2 + 9 x + 23 ( x −1 ) ( x 2 + 6 x + 11 ) 4 x 2 + 9 x + 23 ( x −1 ) ( x 2 + 6 x + 11 )

−2 x 2 + 10 x + 4 ( x −1 ) ( x 2 + 3 x + 8 ) −2 x 2 + 10 x + 4 ( x −1 ) ( x 2 + 3 x + 8 )

x 2 + 3 x + 1 ( x + 1 ) ( x 2 + 5 x −2 ) x 2 + 3 x + 1 ( x + 1 ) ( x 2 + 5 x −2 )

4 x 2 + 17 x −1 ( x + 3 ) ( x 2 + 6 x + 1 ) 4 x 2 + 17 x −1 ( x + 3 ) ( x 2 + 6 x + 1 )

4 x 2 ( x + 5 ) ( x 2 + 7 x −5 ) 4 x 2 ( x + 5 ) ( x 2 + 7 x −5 )

4 x 2 + 5 x + 3 x 3 −1 4 x 2 + 5 x + 3 x 3 −1

−5 x 2 + 18 x −4 x 3 + 8 −5 x 2 + 18 x −4 x 3 + 8

3 x 2 −7 x + 33 x 3 + 27 3 x 2 −7 x + 33 x 3 + 27

x 2 + 2 x + 40 x 3 −125 x 2 + 2 x + 40 x 3 −125

4 x 2 + 4 x + 12 8 x 3 −27 4 x 2 + 4 x + 12 8 x 3 −27

−50 x 2 + 5 x −3 125 x 3 −1 −50 x 2 + 5 x −3 125 x 3 −1

−2 x 3 −30 x 2 + 36 x + 216 x 4 + 216 x −2 x 3 −30 x 2 + 36 x + 216 x 4 + 216 x

For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor.

3 x 3 + 2 x 2 + 14 x + 15 ( x 2 + 4 ) 2 3 x 3 + 2 x 2 + 14 x + 15 ( x 2 + 4 ) 2

x 3 + 6 x 2 + 5 x + 9 ( x 2 + 1 ) 2 x 3 + 6 x 2 + 5 x + 9 ( x 2 + 1 ) 2

x 3 − x 2 + x −1 ( x 2 −3 ) 2 x 3 − x 2 + x −1 ( x 2 −3 ) 2

x 2 + 5 x + 5 ( x + 2 ) 2 x 2 + 5 x + 5 ( x + 2 ) 2

x 3 + 2 x 2 + 4 x ( x 2 + 2 x + 9 ) 2 x 3 + 2 x 2 + 4 x ( x 2 + 2 x + 9 ) 2

x 2 + 25 ( x 2 + 3 x + 25 ) 2 x 2 + 25 ( x 2 + 3 x + 25 ) 2

2 x 3 + 11 x 2 + 7 x + 70 ( 2 x 2 + x + 14 ) 2 2 x 3 + 11 x 2 + 7 x + 70 ( 2 x 2 + x + 14 ) 2

5 x + 2 x ( x 2 + 4 ) 2 5 x + 2 x ( x 2 + 4 ) 2

x 4 + x 3 + 8 x 2 + 6 x + 36 x ( x 2 + 6 ) 2 x 4 + x 3 + 8 x 2 + 6 x + 36 x ( x 2 + 6 ) 2

2 x −9 ( x 2 − x ) 2 2 x −9 ( x 2 − x ) 2

5 x 3 −2 x + 1 ( x 2 + 2 x ) 2 5 x 3 −2 x + 1 ( x 2 + 2 x ) 2

For the following exercises, find the partial fraction expansion.

x 2 + 4 ( x + 1 ) 3 x 2 + 4 ( x + 1 ) 3

x 3 −4 x 2 + 5 x + 4 ( x −2 ) 3 x 3 −4 x 2 + 5 x + 4 ( x −2 ) 3

For the following exercises, perform the operation and then find the partial fraction decomposition.

7 x + 8 + 5 x −2 − x −1 x 2 −6 x −16 7 x + 8 + 5 x −2 − x −1 x 2 −6 x −16

1 x −4 − 3 x + 6 − 2 x + 7 x 2 + 2 x −24 1 x −4 − 3 x + 6 − 2 x + 7 x 2 + 2 x −24

2 x x 2 −16 − 1 −2 x x 2 + 6 x + 8 − x −5 x 2 −4 x 2 x x 2 −16 − 1 −2 x x 2 + 6 x + 8 − x −5 x 2 −4 x

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Partial Fractions

A way of "breaking apart" fractions with polynomials in them.

What are Partial Fractions?

We can do this directly:

2 x−2 + 3 x+1 = 2(x+1) + 3(x−2) (x−2)(x + 1)

Which can be simplified using Rational Expressions to:

= 2x+2 + 3x−6 x 2 +x−2x−2

= 5x−4 x 2 −x−2

... but how do we go in the opposite direction?

That is what we are going to discover:

How to find the "parts" that make the single fraction (the " partial fractions ").

Why Do We Want Them?

First of all ... why do we want them?

Because the partial fractions are each simpler .

This can help solve the more complicated fraction. For example it is very useful in Integral Calculus .

Partial Fraction Decomposition

The method is called "Partial Fraction Decomposition" , and goes like this:

Step 1: Factor the bottom:

Step 2: Write one partial fraction for each of those factors:

Step 3: Multiply through by the bottom so we no longer have fractions:

Step 4: Now find the constants A 1 and A 2 :

Substituting the roots, or "zeros", of (x−2)(x+1) can help:

And we have our answer:

That was easy! ... almost too easy ...

... because it can be a lot harder !

Now we go into detail on each step.

Proper Rational Expressions

Firstly, this only works for Proper Rational Expressions, where the degree of the top is less than the bottom.

The degree is the largest exponent the variable has.

If your expression is Improper, then do polynomial long division first.

Factoring the Bottom

It is up to you to factor the bottom polynomial. See Factoring in Algebra .

But don't factor them into complex numbers ... you may need to stop some factors at quadratic (called irreducible quadratics because any further factoring leads to complex numbers):

Example: (x 2 −4)(x 2 +4)

• x 2 −4 can be factored into (x−2)(x+2)
• But x 2 +4 factors into complex numbers, so don't do it

So the best we can do is:

(x−2)(x+2)(x 2 +4)

So the factors could be a combination of

• linear factors
• irreducible quadratic factors

When you have a quadratic factor you need to include this partial fraction:

B 1 x + C 1 (Your Quadratic)

Factors with Exponents

Sometimes you may get a factor with an exponent, like (x−2) 3 ...

You need a partial fraction for each exponent from 1 up.

1 (x−2) 3

Has partial fractions

A 1 x−2 + A 2 (x−2) 2 + A 3 (x−2) 3

The same thing can also happen to quadratics:

1 (x 2 +2x+3) 2

Has partial fractions:

B 1 x + C 1 x 2 +2x+3 + B 2 x + C 2 (x 2 +2x+3) 2

Sometimes Using Roots Does Not Solve It

Even after using the roots (zeros) of the bottom you can end up with unknown constants.

So the next thing to do is:

Gather all powers of x together and then solve it as a system of linear equations .

Oh my gosh! That is a lot to handle! So, on with an example to help you understand:

A Big Example Bringing It All Together

Here is a nice big example for you!

x 2 +15 (x+3) 2 (x 2 +3)

• Because (x+3) 2 has an exponent of 2, it needs two terms ( A 1 and A 2 ).
• And (x 2 +3) is a quadratic, so it will need Bx + C :

x 2 +15 (x+3) 2 (x 2 +3)   =   A 1 x+3 + A 2 (x+3) 2 + Bx + C x 2 +3

Now multiply through by (x+3) 2 (x 2 +3) :

x 2 +15 = (x+3)(x 2 +3)A 1 + (x 2 +3)A 2 + (x+3) 2 (Bx + C)

There is a zero at x = −3 (because x+3=0), so let us try that:

(−3) 2 +15 = 0 + ( (−3) 2 +3)A 2 + 0

And simplify it to:

Let us replace A 2 with 2 :

x 2 +15 = (x+3)(x 2 +3)A 1 + 2x 2 +6 + (x+3) 2 (Bx + C)

Now expand the whole thing:

x 2 +15 = (x 3 +3x+3x 2 +9)A 1 + 2x 2 +6 + (x 3 +6x 2 +9x)B + (x 2 +6x+9)C

Gather powers of x together:

x 2 +15 = x 3 (A 1 +B)+x 2 (3A 1 +6B+C+2)+x(3A 1 +9B+6C)+(9A 1 +6+9C)

Separate the powers and write as a Systems of Linear Equations :

Simplify, and arrange neatly:

You can choose your own way to solve this ... I decided to subtract the 4th equation from the 2nd to begin with:

Then subtract 2 times the 1st equation from the 2nd:

Now I know that B = −(1/2) .

We are getting somewhere!

And from the 1st equation I can figure that A 1 = +(1/2) .

And from the 4th equation I can figure that C = +(1/2) .

Final Result:

And we can now write our partial fractions:

x 2 +15 (x+3) 2 (x 2 +3)   =   1 2(x+3) + 2 (x+3) 2 + −x + 1 2(x 2 +3)

Phew! Lots of work. But it can be done.

(Side note: It took me nearly an hour to do this, because I had to fix 2 silly mistakes along the way!)

• Start with a Proper Rational Expressions (if not, do division first)
• or "irreducible" quadratic factors
• Write out a partial fraction for each factor (and every exponent of each)
• Multiply the whole equation by the bottom
• substituting zeros of the bottom
• making a system of linear equations (of each power) and solving
• Write out your answer!

Partial Fraction Decomposition

How to perform partial fraction decomposition or expansion.

This method is used to decompose a given rational expression into simpler fractions. In other words, if I am given a single complicated fraction, my goal is to break it down into a series of “smaller” components or parts.

Previously on adding/subtracting rational expressions, we want to combine two or more rational expressions into a single fraction just like the example below. However, partial fraction decomposition ( also known as partial fraction expansion ) is precisely the reverse process of that. The following is an illustrative diagram to show the main concept.

 Now, I will go over five (5) examples to demonstrate the steps involved in decomposing a single fraction into parts.

Examples of How to Decompose Partial Fractions

Example 1: Find the partial fraction decomposition of the rational expression.

This problem is easy, so think of this as an introductory example. I will start by factoring the denominator (take out [latex]x[/latex] from the binomial). Next, I will set up the decomposition process by placing [latex]A[/latex] and [latex]B[/latex] for each of the unique or distinct linear factors. The subsequent steps then involve getting rid of all the denominators by multiplying the [latex]LCD[/latex] (which is just the original denominator of the problem) throughout the entire equation.

I should end up with a simple equation where I can easily compare the coefficients of similar terms from both sides of the equation. As a result, I will get a system of linear equations with variables [latex]A[/latex] and [latex]B[/latex] that can be solved by either the Substitution Method or Elimination Method , whichever I prefer.

• Given the fraction
• Factor out the denominator.
• Create individual fractions on the right side having each of the factors acting as the denominator. I have two partial fractions here with two unknown values of numerators represented by variables [latex]A[/latex] and [latex]B[/latex].
• I want to eliminate all the denominators. It can be done by multiplying both sides of the equation by the [latex]\color{blue}LCD = x\left( {x + 1} \right)[/latex].
• Next, I distribute the [latex]LCD[/latex] to each side of the equation. At this stage, I try to be very careful in keeping track of the cancellation process. I want to make sure that I get this step right to prevent any unnecessary headaches later.
• This is the simplified equation after doing the above step correctly.
• Now, I will multiply things out and collect common terms by writing them side by side. It’s time to compare the coefficients of the polynomials. The idea is to equate the corresponding coefficients of similar terms.
• I equate the coefficients of the [latex]x-[/latex]term emphasized by the yellow highlight. In addition, I equate the constant terms as shown by the green highlight.
• I then arrive at solving two equations with two unknowns. Use the Substitution Method to solve for [latex]B[/latex].
• These are the final values of variables [latex]A[/latex] and [latex]B[/latex].
• Plug in the solved values of [latex]A[/latex] and [latex]B[/latex] back into the original setup to get the final answer.

Example 2: Find the partial fraction decomposition of the rational expression.

This problem is similar to example 1. The only difference is that the factors of the denominator are two linear binomials.

• Given the problem
• I start by factoring out the trinomial in the denominator.

Then, I setup up the partial fraction decomposition by putting [latex]A[/latex] and [latex]B[/latex] as numerators. The two distinct linear factors will take the position of the denominators.

• I want to eliminate all the denominators so I will multiply both sides of the equation by the [latex]LCD[/latex] (in blue).
• I need to be careful when canceling common factors.
• The steps here are pretty much part of the “cleaning up” process and reorganization of common terms.
• It’s time to create the correspondence between the two sides of the equation.

For the [latex]x[/latex] terms, the coefficients [latex]\left( {A + B} \right) = 1[/latex] while the pure numbers I have [latex]3A + 5B = – 1[/latex].

• We have two equations with two unknowns. From this point forward, it should be easy to handle, right? I suggest we use the Elimination Method to solve for [latex]A[/latex] and [latex]B[/latex].

Multiply the top equation by either [latex] – \,3[/latex] or [latex] – \,5[/latex] then add them together to eliminate one of the variables.

• These are the correct values of [latex]A[/latex] and [latex]B[/latex].
• I will come back to the original setup of the partial fractions to replace the values of [latex]A[/latex] and [latex]B[/latex] with actual numbers.

Example 3: Find the partial fraction decomposition of the rational expression

In this problem, the denominator is the product of a distinct linear factor which is repeated three times denoted by the exponent [latex]3[/latex]. Don’t commit the error of writing three partial fractions with a common denominator of just [latex]\left( {x – 1} \right)[/latex]. That is not correct.

Instead, think of three possibilities on how the denominator may look like after solving it. Possible denominators include [latex]\left( {x – 1} \right)[/latex], [latex]{\left( {x – 1} \right)^2}[/latex], and [latex]{\left( {x – 1} \right)^3}[/latex].

• Since I have a repeated linear factor [latex]\left( {x – 1} \right)[/latex] raised to the power of [latex]3[/latex], I need to account for each power starting from lowest (1) to highest (3).

Do you see why this setup is wrong?

Hint: Add the partial fractions then compare their denominators.

• Remove all the denominators by distributing the [latex]LCD[/latex] into the equation.
• This is the simplified version after the cancellations of common factors.
• Like in our previous problem, I will distribute [latex]A[/latex] and [latex]B[/latex] into the parenthesis. Then, rearrange them in such a way that similar terms are side-by-side.

Finally, I’ll group the [latex]x^2[/latex], [latex]x[/latex], and “constant” coefficients together.

• Use arrows, if necessary, to create correspondence among coefficients of similar terms.

I introduced a zero placeholder to the missing [latex]x^2[/latex] term on the left side. This is a very critical step.

• Now, I can clearly extract the required equations that will be used to solve for the missing variables.

The equation [latex]A = 0[/latex] is a nice freebie since I don’t have to sweat solving for it algebraically.

• Use the fact that [latex]A = 0[/latex], plug that to the second equation [latex]\left( { – 2A + B = 5} \right)[/latex] to get [latex]B[/latex]. Finally, solve for [latex]C[/latex] using the third equation using the solved values of [latex]A[/latex] and [latex]B[/latex].

Did you get the same answers?

• Write down the original setup of partial fraction decomposition, and replace the solved values for [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex].

The fraction where the numerator is [latex]A = 0[/latex] will disappear. This leaves us with two fractions as the final answer.

Example 4: Find the partial fraction decomposition of the rational expression

This is the case where the denominator is a product of distinct linear factors where some are repeated.

Notice that the denominator of this rational expression is composed of two distinct linear factors. The first one is [latex]\left( {x – 2} \right)[/latex] which appears once, while the second factor is [latex]\left( {x – 3} \right)[/latex] appears twice, thus repeated.

• I have two distinct linear factors here.

[latex]\left( {x – 2} \right)[/latex] appears once

[latex]\left( {x – 3} \right)[/latex] appears twice denoted by the power [latex]2[/latex]

Therefore, I will write [latex]\left( {x – 2} \right)[/latex] in one partial faction, while [latex]\left( {x – 3} \right)[/latex] in two partial fractions with increasing exponents from [latex]1[/latex] to [latex]2[/latex].

• I will eliminate all denominators by multiplying the equation with the respective [latex]LCD[/latex].
• Again, I have to be careful with my cancellations.
• You might say that this looks “horrible” at the surface. However, if you look at each step patiently, you should realize that it’s not that bad.

I simply distribute [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex] into the parenthesis. Then, rearrange them so that similar terms are adjacent to each other.

Finally, I’ll group the coefficients of [latex]x^2[/latex], [latex]x[/latex], and “constant” terms together.

• Only the left side has the [latex]x^2[/latex] term. This means I must provide zero placeholders for the missing [latex]x[/latex] term and the constant term.

By doing so, it becomes easy to compare the coefficients of similar terms on both sides of the equation.

The arrows and color coding should guide you with this process.

• I can solve this in several ways. One way is to use the Elimination Method to get rid of [latex]C[/latex] between the 2nd and 3rd equations.

I will multiply the 2nd equation by [latex]2[/latex] then add that to the 3rd equation. I should end up with an equation containing [latex]A[/latex] and [latex]B[/latex] only. Use this “new” equation with the 1st equation to solve for the values of [latex]A[/latex] and [latex]B[/latex]. I suggest that you try it on paper so you can follow it.

Once you get the values of [latex]A[/latex] and [latex]B[/latex], you can solve for [latex]C[/latex] using either of the 2nd or 3rd equation using back-substitution.

• These are the correct values of [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex].
• Substitute the values into the original partial fraction setup to obtain the final answer.

Example 5: Find the partial fraction decomposition of the rational expression

This is another type of problem in partial fraction decomposition. Factoring the denominator, I get the following.

Notice that this time around I have a quadratic factor. The question is, can I still factor it out further into linear terms? I can try, but it’s obvious that it can’t be factored out anymore. This, in fact, has a special name called irreducible quadratic .

This problem, therefore, is a case where the denominator is a product of a distinct linear factor and an irreducible quadratic factor which are both non-repeating.

• Two kinds of factors here.

Linear factor: [latex]x[/latex]

Irreducible quadratic: [latex]{x^2}-2[/latex]

Notice that the degree of the numerator is always one less than the degree of the denominator.

The numerator of the linear factor [latex]x[/latex] is a constant [latex]A[/latex].

The numerator of the quadratic factor [latex]{x^2}-2[/latex] is a linear term [latex]Bx + C[/latex].

• Get rid of the denominators by multiplying both sides by the [latex]LCD[/latex].
• Cancel out the common factors to clean it up!
• I will distribute, rearrange, and group the coefficients of similar terms.
• Provide a zero placeholder for the [latex]x^2[/latex] on the left side.

Observe the correspondence of the coefficients on both sides of the equation.

• By quick analysis, I know that [latex]C = 1[/latex] and [latex]A = -2[/latex].

Since [latex]A+B=0[/latex] and [latex]A = -2[/latex], therefore [latex]\left( { – 2} \right) + B = 0[/latex] implies [latex]B = 2[/latex].

• Substitute back the values into the original setup of the partial fraction decomposition, and we’re done!

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Partial Fractions

Introduction to partial fractions, learning objectives.

By the end of this section, you will be able to:

• Decompose   [latex]\frac{{P( x )}}{{ Q( x )}}[/latex] ,  where  Q( x )  has only nonrepeated linear factors.
• Decompose  [latex]\frac{{P( x )}}{{ Q( x )}}[/latex] ,  where  Q( x )  has repeated linear factors.
• Decompose  [latex]\frac{{P( x )}}{{ Q( x )}}[/latex] ,  where  Q( x )  has a nonrepeated irreducible quadratic factor.
• Decompose  [latex]\frac{{P( x )}}{{ Q( x )}}[/latex] ,  where  Q( x )  has a repeated irreducible quadratic factor.

Earlier in this chapter, we studied systems of two equations in two variables, systems of three equations in three variables, and nonlinear systems. Here we introduce another way that systems of equations can be utilized—the decomposition of rational expressions.

Fractions can be complicated; adding a variable in the denominator makes them even more so. The methods studied in this section will help simplify the concept of a rational expression.

• Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution

AP Calculus BC Review: Partial Fractions

The Method of Partial Fractions is actually a technique of algebra that allows you to rewrite certain kinds of rational expressions in a more useful way.

In this review, we will discuss the how and when to use the method in integral problems, especially those found on the AP Calculus BC exam.

The Method of Partial Fractions (PF)

The method is actually the reverse of adding rational expressions.

Suppose you have a rational function , that is, a fractional expression of two polynomial.

The question is this: Can we find two or more simpler rational expressions that add to the given one?

The key is to factor the denominator. The denominators of the new fractions will involve these factors. Then, working backward, you can figure out what the new numerators must be in order to arrive at your given function.

Of course, there are many details that I’m leaving out at the moment. We’ll explore the method in more detail below.

PF for Non-Repeating Linear Factors

First, PF only works when the degree of the denominator is greater than the degree of the numerator. If the numerator has higher degree than the denominator, then you must first do polynomial (or synthetic ) division . I have yet to see a problem on the Calculus BC exam that requires polynomial division, so we’ll skip over that in this review.

Furthermore, on the AP Calculus BC exam, you will only need to know about denominators that factor into non-repeating linear factors.

In other words, the denominator will factor completely into unique factors:

The next step after factoring is to write down as many fractions as you have factors. We know what the denominators will be (those are the factors themselves). But we don’t know the numerators yet. For now, just write the numerators as variables. We’ll have to solve for them later.

Next you’ll need to figure out exactly what those numbers are on top, A 1 , …, A n . By the way, when there are only two or three factors, I usually call these numbers A , B , C , etc. The actual variable name doesn’t matter.

Although there is no calculus involved at this step, this is where a lot of students seem to get stuck. The algebra can be very tricky and tedious. So I’ll explain it using two different methods on the same example.

Finally, integrate each term separately. When the factors are fractions with a linear denominator, you can expect the antiderivative to involve natural logarithms.

Example Using Partial Fractions

Decomposing the expression.

First, check the degree of the top and bottom. Since the denominator has degree 2 while the numerator has degree 1, we may use PF.

The method always begins with factoring the denominator. Sometimes this step is already done in a given problem on the BC exam, but not always.

Now since there are two factors, there will be two separate fractions. Let’s let A and B stand for the unknown numerators.

From here, there are two ways to go.

Method #1 — Recombine Fractions

The idea is that we want the sum of the fractions to equal the original rational expression. So use your algebra skills to recombine the fractions and compare the results.

You’ll need a common denominator, but that’s the easy part! You just have to multiply the two denominators back together.

So, after grouping like terms, we set the numerators equal to one another. Comparing the x -terms and constant terms, this leads to a system of two equations.

Now there is a variety of ways to solve such systems. Matrix methods work, if you know how to use them.

Alternatively, you could do simple substitution.

Solve the first equation for A , to get: A = 5 – 3 B . Then plug in that expression for A in the second equation and solve for B .

Finally solve for A by plugging in the known B -value. A = 5 – 3(3) = -4.

Therefore, we now have a partial fractions decomposition for the original rational expression.

Surely this is the end of the process right? Well no, you still have to integrate the decomposed form. (We’ll do that eventually.)

But before getting to that, let’s discuss other methods for decomposing the fraction which is more efficient than this one.

Method #2 — Heaviside Cover-Up Method

The Heaviside Cover-Up Method avoids all that algebra, but seems more like a “trick.” It’s guaranteed to work for non-repeated linear factors, but the reason it works is kind of subtle.

Here’s how you do it.

For each factor, first find it’s root , that is, the value of x that results when you set it equal to zero and solve.

Then cover up that factor in the original rational expression, and plug the root into x everywhere else you see it in the expression.

Evaluate it, and this will be the A i (numerator) value in the PF decomposition.

Going back to our example, remember we factored and decomposed the function so that:

The first factor is 3 x + 1. Setting it equal to zero and solving, we get:

Now we plug this value back into the original expression, but ignore the factor (3x + 1).

This tells us that A = -4.

Then move on to the second factor, ( x – 2), whose root is x = 2. Ignore that factor now and plug in.

Now we have B = 3.

The end result is the same as with Method #1, but it takes far less time.

Final Step: Integrate!

After completing either the algebraic method or Heaviside Cover-Up, you would now be in position to find the antiderivative.

Just be careful: There are usually simple u -substitutions that must be made in order to get the right answers.

Though at first it may seem as though Partial Fractions is difficult, it’s more like the challenge of learning to ride a bike.

At first you don’t know what you’re doing and you might fall down a lot. But with enough practice, and especially the willingness to get back on the bike after falling off, then you’ve learned a skill that will stay with you for the rest of your life!

In fact, similar to riding a bike, I think that PF is kinda fun. (But my close friends assure me that I’m just weird.)

Check out the following resource for other important integration technques: AP Calculus Review: Indefinite Integrals .

Shaun earned his Ph. D. in mathematics from The Ohio State University in 2008 (Go Bucks!!). He received his BA in Mathematics with a minor in computer science from Oberlin College in 2002. In addition, Shaun earned a B. Mus. from the Oberlin Conservatory in the same year, with a major in music composition. Shaun still loves music — almost as much as math! — and he (thinks he) can play piano, guitar, and bass. Shaun has taught and tutored students in mathematics for about a decade, and hopes his experience can help you to succeed!

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A Level Maths

Maths A-Level Resources for AQA, OCR and Edexcel

Partial Fractions

Remember these formulas of partial fractions for different types of fractions:

We can recall from GCSE’s that to transform a function consisting of many fractions into a single fraction, we take LCM (lowest common factor) of the entire function i.e:

Now the question is what do we do when we want to reverse this process and split a single fraction into two or more fractions. Well, the process of breaking a single fraction into multiple fractions is known as splitting into ”partial fractions” . It could be both sum or difference of two or more fractions.

There are three different types of fractions:

1.     Where a fraction consists of only linear factors in the denominator.

2.     Where there are repeated factors in the denominator of the fraction.

3.     Where there are quadratic factors in the denominator of the fraction.

We will go through each one of the types with the methods used to solve them along with examples below.

1.     Linear factors in the denominator

This included both proper fractions and improper fractions. Let’s have a look at the proper fractions first.

Note: This is the same function that resulted by taking LCM of fractions in the beginning of this article.

Since the denominator has linear factors, there required partial fractions will be:

First find the 2 values of x :

Substitute each value of x in equation 1, one at a time.

So to find the value of A put x = -1 in equation 1,

Substituting the values of A and B in equation (i) above gives us our partial fractions:

For such proper fractions whose denominators are linear factors we can also use a cover up method.

Cover up method

This is basically a shortcut of finding the partial fractions, where we don’t have to do long calculations like we did in the above example i.e let’s do the above example now with the cover up method. You will see how quickly we can find the results.

As we know the partial fraction expression would be:

To find A we consider the left hand side of the equation:

We cover up one factor in the denominator first i.e cover up (x + 1)

Since we have covered up (x + 1) , the value of x in this case is -1.

Substituting this value of x we get:

Now that we have understood how we find partial fractions for proper fractions, we move on to improper fractions.

Partial fractions of Improper fractions

Improper fractions are fractions whose degree of denominator is equal to or less than the degree of its numerator i.e:

these are both considered as improper fractions.

To find work out the partial fractions, we must have the function as a proper fraction. Therefore, we convert all improper fractions into proper ones before we decompose them into partial fractions. We do this by dividing the numerator by its denominator till it becomes a proper fractions. This is done through algebraic long division. Algebraic long division has been explained in detail in the article ”Algebraic long division”. Let’s work out an example now.

We can see that the above function is an improper fractions as the degree of numerator is equal to degree of the denominator. Hence, we must carry out long division to convert it into a proper fraction.

After the long division the fraction becomes:

Now using the cover up method we find the values of A and B.

Therefore, the required partial fractions are:

2.    Repeated factors in the denominator

When x = 1:

When x = 2:

To find B simplify eq 1:

0 = A + B  we know  A = 1

B =  -1

Hence, the required partial fractions are:

3.    Quadratic factors in the denominator

In the case, where a fraction has a quadratic factor in the denominator which cannot be simplified further, then that denominator will have a linear numerato in its partial fraction i.e:

We will then follow the same process as above to find the values of A and after that we compare the coefficients of x to find the value of B and C .

• https://www.toppr.com/guides/maths/integrals/integration-by-partial-fractions/
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• partial\:fractions\:\frac{x}{(x+1)(x-4)}
• partial\:fractions\:\frac{(2(1+x))}{x(x^{2}+3x+2)}
• partial\:fractions\:\frac{1}{(x+5)(x+4)}
• partial\:fractions\:\frac{x^{2}+2}{(x+2)^{2}(x+3)}
• What is partial fraction?
• Partial fractions is a technique used in algebra to decompose a rational function into simpler fractions.
• How do you solve partial fractions?
• To solve partial fractions, you first factor the denominator of the rational function into linear or quadratic factors. Then, you express the original function as a sum of simpler fractions with denominators equal to these factors, and unknown numerators which can be determined by comparing coefficients.
• What are the 4 types of partial fractions?
• The 4 types of partial fractions are Linear factors with distinct roots, Linear factors with repeated roots, Quadratic factors with distinct roots and Quadratic factors with repeated roots
• What is a Linear partial fraction?
• A linear partial fraction is a partial fraction in which the denominator factors into linear factors. In other words, the denominator of the rational function is a product of expressions of the form (ax + b), where a and b are constants.
• What is a Quadratic partial fraction?
• A quadratic partial fraction is a partial fraction in which the denominator factors into quadratic factors. In other words, the denominator of the rational function is a product of expressions of the form (ax^2+bx + c), where a, b and c are constants.
• What is a Repeated linear partial fraction?
• A repeated linear partial fraction is a partial fraction in which the denominator has repeated linear factors. In other words, the denominator of the rational function is a product of expressions of the form (ax + b)^n, where a and b are constants, and n is a positive integer greater than 1.
• What is a General partial fraction?
• A general partial fraction is a partial fraction that includes all possible types of factors in the denominator of a rational function, including linear factors with distinct roots, linear factors with repeated roots, quadratic factors with distinct roots, and quadratic factors with repeated roots.

partial-fractions-calculator

• High School Math Solutions – Partial Fractions Calculator Partial fractions decomposition is the opposite of adding fractions, we are trying to break a rational expression...
• Partial Fractions Learning Objectives By the end of this section, you will be able to: Decompose , where Q( x ) has only nonrepeated linear factors. Decompose , where Q( x ) has repeated linear factors. Decompose , where Q( x ) has a nonrepeated irreducible quadratic factor. Decompose , where Q( x ) has a repeated i...
• Partial Fractions: an Application of Systems Earlier in this chapter, we studied systems of two equations in two variables, systems of three equations in three variables, and nonlinear systems. Here we introduce another way that systems of equations can be utilized—the decomposition of rational expressions. Fractions can be complicated; adding...

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• Math Article

Partial Fractions

Partial fractions are the fractions used for the decomposition of a rational expression. When an algebraic expression is split into a sum of two or more rational expressions, then each part is called a partial fraction. Hence, basically, it is the reverse of the addition of rational expressions. Similar to fractions , a partial fraction will have a numerator and denominator, where the denominator represents the decomposed part of a rational function.

In mathematics, we can see many complex rational expressions. If we try to solve the problems in a complex form, it will take a lot of time to find the solution. To avoid this complexity, we have to continue the problem by reducing the complex form of the rational expression into the simpler form. Partial fraction decomposition is one of the methods, which is used to decompose rational expressions into simpler partial fractions. This process is more useful in the integration process. In this article, you will learn the definition of the partial fraction, partial fraction decomposition, partial fractions of an improper fraction with solved examples in detail.

What is a Partial Fraction?

An algebraic fraction can be broken down into simpler parts known as “ partial fractions “. Consider an algebraic fraction, (3x+5)/(2x 2 -5x-3). This expression can be split into simple form like [2/(x – 3)] – [1/(2x + 1)].

The simpler parts  [2/(x – 3)] and [1/(2x + 1)]  are known as partial fractions.

This means that the algebraic expression can be written  in the form, as given in the figure :

Note: The partial fraction decomposition only works for the proper rational expression (the degree of the numerator is less than the degree of the denominator). In case, if the rational expression is an improper rational expression (the degree of the numerator is greater than the degree of the denominator), first do the division operation to convert it into proper rational expression. This can be achieved with the help of a  polynomial long division method .

Partial Fraction Formulas

Here the list of Partial fractions formulas is given. These formulas will help us to decompose a rational expression into partial fractions. These are common types of partial fractions which are used to solve problems.

Here A, B and C are real numbers.

Partial Fractions of Rational Functions

Any number which can be easily represented in the form of p/q, such that p and q are integers and q≠0 is known as a rational number. Similarly, we can define a rational function as the ratio of two polynomial functions P(x) and Q(x), where P and Q are polynomials in x and Q(x)≠0. A rational function is known as proper if the degree of P(x) is less than the degree of Q(x); otherwise, it is known as an improper rational function. With the help of the long division process, we can reduce improper rational functions to proper rational functions. Therefore, if P(x)/Q(x) is improper, then it can be expressed as:

\(\begin{array}{l}\frac{P(x)}{Q(x)}= A(x) + \frac{R(x)}{Q(x)}\end{array} \)

Here, A(x) is a polynomial in x and R(x)/Q(x) is a proper rational function.

We know that the integration of a function f(x) is given by F(x) and it is represented by:

∫f(x)dx = F(x) + C

Here R.H.S. of the equation means integral of f(x) with respect to x and C is the constant of integration .

Decomposition of Partial Fractions

In order to integrate a rational function, it is reduced to a proper rational function. The method in which the integrand is expressed as the sum of simpler rational functions is known as decomposition into partial fractions . After splitting the integrand into partial fractions, it is integrated accordingly with the help of traditional integrating techniques.

The stepwise procedure for finding the partial fraction decomposition is explained here::

• Step 1: While decomposing the rational expression into the partial fraction, begin with the proper rational expression.
• Step 2: Now, factor the denominator of the rational expression into the linear factor or in the form of irreducible quadratic factors (Note: Don’t factor the denominators into the complex numbers).
• Step 3: Write down the partial fraction for each factor obtained, with the variables in the numerators, say A and B.
• Step 4: To find the variable values of A and B, multiply the whole equation by the denominator.
• Step 5: Solve for the variables by substituting zero in the factor variable.
• Step 6: Finally, substitute the values of A and B in the partial fractions .

Hence, the expression is decomposed into partial fractions.

Partial Fraction of Improper Fraction

An algebraic fraction is improper if the degree of the numerator is greater than or equal to that of the denominator. The degree is the highest power of the polynomial. Suppose, m is the degree of the denominator and n is the degree of the numerator. Then, in addition to the partial fractions arising from factors in the denominator, we must include an additional term: this additional term is a polynomial of degree n − m.

• A polynomial with a zero degree is K, where K is a constant
• A polynomial of degree 1 is Px + Q
• A polynomial of degree 2 is Px 2 +Qx+K

Video Lesson on Improper Fractions

Partial Fraction in Integration

Let us look into an example to have a better insight into integration using partial fractions.

Example: Integrate the function \(\begin{array}{l}\frac{1}{(x-3)(x+1)}\end{array} \) with respect to x.

Solution: The given integrand can be expressed in the form of partial fraction as:

\(\begin{array}{l}\frac{1}{(x-3)(x+1)} = \frac{A}{(x-3)} + \frac{B}{(x+1)}\end{array} \)

To determine the value of real coefficients A and B, the above equation is rewritten as:

1= A(x+1)+B(x-3)

⇒1=x(A+B)+A-3B

Equating the coefficients of x and the constant, we have

Solving these equations simultaneously, the value of A =1/4 and B = -1/4. Substituting these values in equation 1, we have

\(\begin{array}{l}\frac{1}{(x-3)(x+1)} = \frac{1}{4(x-3)} + \frac{-1}{4(x+1)}\end{array} \)

Integrating with respect to x we have;

\(\begin{array}{l}\int \frac{1}{(x-3)(x+1)} = \int \frac{1}{4(x-3)} + \int \frac{-1}{4(x+1)}\end{array} \)

According to the properties of integration, the integral of the sum of two functions is equal to the sum of integrals of the given functions, i.e.,

∫[f(x) +g(x)]dx = ∫f(x)dx + ∫g(x)dx

\(\begin{array}{l}= \frac{1}{4} \int \frac{1}{(x-3)} – \frac{1}{4} \int \frac{1}{(x+1)}\end{array} \)

\(\begin{array}{l}= \frac{1}{4} \ln \left | x-3 \right | – \frac{1}{4} \ln \left | x+1 \right |\end{array} \)

\(\begin{array}{l}= \frac{1}{4} \ln \left | \frac{x-3}{x+1} \right |\end{array} \)

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Partial Fractions – Solved Examples

Example 1: Write the partial fraction decomposition of the following expression.

(20x + 35)/(x + 4) 2

(20x + 35)/(x + 4) 2   = [A/(x + 4)] + [B/(x + 4) 2 ]

(20x + 35)/(x + 4) 2   = [A(x + 4) + B]/ (x + 4) 2

Now, equating the numerators,

20x + 35 = A(x + 4) + B

20x + 35 = Ax + 4A + B

20x + 35 = Ax + (4A + B)

By equating the coefficients,

4A + B = 35

4(20) + B = 35

B = 35 – 80 = -45

Therefore, (20x + 35)/(x + 4) 2  = [20/(x + 4)] – [45/(x + 4) 2 ]

Example 2: Decompose the given expression into partial fractions.

(x 2  + 1)/ (x 3  + 3x 2  + 3x + 2)

Using the factor theorem, x + 2 is a factor of x 3  + 3x 2  + 3x + 2.

Thus, x 3  + 3x 2  + 3x + 2 = (x + 2)(x 2  + x + 1)

Now, the given expression can be written as:

(x 2  + 1)/ (x 3  + 3x 2  + 3x + 2) = (x 2  + 1)/ [(x + 2)(x 2  + x + 1)]

By the method of decomposition,

(x 2  + 1)/(x + 2)(x 2  + x + 1) = [A/(x + 2)] + [(Bx + C)/(x 2  + x + 1)]

(x 2  + 1)/(x + 2)(x 2 + x + 1) = [A(x 2  + x + 1) + (Bx + C)(x + 2)]/ [(x + 2)(x 2  + x + 1)]

= [(A + B)x 2  + (A + 2B + C)x + A + 2C]/ [(x + 2)(x 2  + x + 1)]

Equating the coefficients in the numerators of both LHS and RHS,

A + 2B + C = 0

Solving these equations,

A = 5/3, B = -2/3 and C = -1/3

(x 2  + 1)/(x + 2)(x 2  + x + 1) = [5/3(x + 2)] – [(2x + 1)/3(x 2  + x + 1)]

Practice Problems

Evaluate the following using the method of partial fractions.

• \(\begin{array}{l}\frac{3x}{(x – 1)(x + 2)}\end{array} \)
• \(\begin{array}{l}\frac{9x^2+5x-3}{(x + 1)^2(x – 2)}\end{array} \)
• \(\begin{array}{l}\frac{x^2 + 2x – 1}{x(x^2 – 1)}\end{array} \)

Frequently Asked Questions (FAQs) on Partial Fractions

What is meant by partial fractions.

In mathematics, the partial fraction is defined as the process of decomposition of a fraction into the simplest form of the fraction.

Write down the procedure for partial fraction decomposition.

The procedure for the partial fraction decomposition is as follows: In a given rational expression, factor the denominator into the linear factors For each factor obtained, write down the partial fraction with variables in the numerator, say x and y To remove the fraction, multiply the whole equation by the denominator factor. Now, solve for the constants x, and y Substitute the constant values in the numerators of the partial fraction, and you will get the solution.

What are the different denominator types in partial fractions?

The four different types of denominators found in the partial fractions are: Linear factors Repeated linear factors Irreducible factors of degree 2 Repeated irreducible factors of degree 2

What is the use of partial fraction decomposition?

Partial fraction decomposition is used to find the inverse Laplace transformation, and also it helps to integrate the rational functions.

What is meant by proper and improper rational expressions?

In proper rational expression, the degree of the numerator is less than the degree of the denominator. Whereas in improper rational expression, the degree of the numerator is greater than the degree of the denominator.

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