: At a Glance

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Topics At a Glance
Indefinite Integrals Introduction
Integration by Substitution: Indefinite Integrals
Legrange (Prime) Notation
Leibniz (Fraction) Notation
Integration by Substitution: Definite Integrals
Integration by Parts: Indefinite Integrals
Some Tricks
Integration by Parts: Definite Integrals
Integration by Partial Fractions
Integrating Definite Integrals
Choosing an Integration Method
Integration by Substitution
Integration by Parts
Integration by Partial Fractions
Thinking Backwards
Improper Integrals
Badly Behaved Limits
Badly Behaved Functions
Badly Behaved Everything
Comparing Improper Integrals
The p-Test
Finite and Infinite Areas
Comparison with Formulas
In the Real World
I Like Abstract Stuff; Why Should I Care?
How to Solve a Math Problem

Introduction to :

Integration by partial fractions is a technique we can use to integrate rational functions when the degree of the numerator is less than the degree of the denominator. Here's the big picture:

We start out with an integral whose integrand is a rational function, like

The degree of the numerator must be less than the degree of the denominator.
We do some sneaky stuff to rewrite the original rational function as a sum of partial fractions:
We integrate the partial fractions, whose antiderivatives all involve the natural log:

Be Careful: When x occurs in a denominator with a coefficient other than 1, you have to use integration by substitution.

Practice:

Examples
Exercises

Example 1

Decompose

into a sum of the form

We want to write

as a sum of fractions

We don't know what A and B are, but we know if we add

and

we'll get

Writing this as an equation, we want to find A and B so that

We know how to add rational functions, so we know that

We want this sum to equal the original fraction, so we want

The denominators of these fractions are the same, so the fractions are the same if their numerators are equal.

This means we need to find A and B so that

A(x + 5) + B(3x + 2) = 7x + 9.

There's an easy way and a harder way to do this. The harder way is to multiply out the left-hand side, find that

(A + 3B)x + (5A + 2B) = 7x + 9,

and solve the system of equations

A + 3B = 7

5A + 2B = 9.

We don't want to do that, because it's too much work. Instead, we'll carefully choose values of x so that we're solving for only one of A or B at a time.

If we take x = -5, for example, then we don't have to worry about A.

A(x + 5) + B(3x + 2) = 7x + 9

A((-5) + 5) + B(3(-5) + 2) = 7(-5) + 9

A(0) + B(-13) = -26

B = 2.

Now that we know B = 2, we can use any value of x we like to solve for A. Let's take x = 0.

A(x + 5) + B(3x + 2) = 7x + 9

A((0) + 5) + (2)(3(0) + 2) = 7(0) + 9

5A + 4 = 9

5A = 5

A = 1.

Now we know that A = 1 and B = 2, so we've successfully decomposed the original fraction:

As long as the denominator of the original function factors into distinct linear factors, there will be one partial fraction for each factor of the denominator. For the time being
we'll be nice and give you the original function with the denominator already factored.

Example 2

Decompose

into partial fractions.

There will be one partial fraction for each factor of the denominator, so we want to find A and B such that

Adding the partial fractions, we get

These fractions have the same denominator, so they will be equal as long as their numerators are equal. This means we need to find A and B such that

7x + 5 = A(x – 1) + B(x + 2).

We want to find A and B the easy way. Choose x = 1 so that we can find B without worrying about A:

7x + 5 = A(x – 1) + B(x + 2)

7(1) + 5 = A((1)-1) + B(1 + 2)

12 = 3B

B = 4.

Now that we know B, we can take any value of x we like and find A. Take x = 0:

7x + 5 = A(x – 1) + B(x + 2)

7(0) + 5 = A((0)-1) + (4)((0) + 2)

5 = -A + 8

A = 3

We've successfully decomposed our original function into partial fractions:

Be Careful: Make sure you put A and B over the correct denominators. It doesn't matter whether we set up the problem as

initially. However, it does matter that we give the answer

and not the answer

One nice thing about partial fractions is that it's easy to check your work, because you know how to add rational functions. We can quickly see that

which does not equal

So if you do accidentally switch the numerators, at least you have a way to find that out.

Example 3

Decompose into partial fractions.

This is the same as all the other problems we just finished doing, with one extra step: we have to factor the denominator to find the denominators of the partial fractions.
Since

x² + 8x + 7 = (x + 7)(x + 1),

we know the partial fractions will be

and

From here, we do the same stuff we did before. We want to have

We add the partial fractions and set the numerator equal to the original numerator:

5x + 23 = A(x + 1) + B(x + 7).

We set x = -1 and solve for B:

5x + 23 = A(x + 1) + B(x + 7)

5(-1) + 23 = A((-1) + 1) + B((-1) + 7)

18 = 6B

3 = B

We set x = 0 and solve for A:

5x + 23 = A(x + 1) + B(x + 7)

5(0) + 23 = A((0) + 1) + (3)((0) + 7)

23 = A + 21

2 = A

Our decomposition is

Example 4

Find

We already found that

Example 5

Find

given that

We know that

To integrate each fraction we have to use substitution, with u = 2x + 3 and u = 5 – x respectively. We get

Previous Example Next Example

Exercise 1

Without a calculator, find

Answer

First we have to put the fractions over a common denominator by multiplying each fraction by a clever form of 1. After the fractions have the same denominator, we can add their numerators.

If there was any part of that problem that gave you trouble, go review fraction addition now. If it was a piece of cake, go ahead and try the next one.

Exercise 2

Find

Answer

As with the numerical fractions, we have to put the rational functions over a common denominator by multiplying each by a clever form of 1. Once the denominators are the same, we can add the numerators.

We could continue to the next step, which would be

but we're not going to bother. We have the right form for the partial fractions technique when we get to the step

so we'll stop there.

When adding rational functions for the sake of the partial fractions technique, we don't multiply things out. For example, when adding

we stop here:

We also don't bother to write out the step where we multiply each fraction by a clever form of 1. We just multiply the numerator of each fraction by the denominator of the other fraction, then add the results.

If this doesn't make sense, we recommend reviewing how to add rational functions. If this does make sense, it's time to practice.

Exercise 3

Find the sum. A and B are unknown numbers.

Answer

Exercise 4

Decompose into partial fractions.

Answer

We need to find A and B so that

Adding the partial fractions,

We must have the numerators equal, so we need to find A and B such that

10x + 27 = A(x + 3) + B(x + 2).

Take x = -3 so we can find B without worrying about A:

10x + 27 = A(x + 3) + B(x + 2)

10(-3) + 27 = A((-3) + 3) + B((-3) + 2)

-3 = -B

B = 3.

Now take x = 0 and find A:

10x + 27 = A(x + 3) + B(x + 2)

10(0) + 27 = A((0) + 3) + (3)((0) + 2)

27 = 3A + 6

7 = A

The final decomposition is

Exercise 5

Decompose into partial fractions.

Answer

We need to find A and B so that

This means we want

and so the numerators must be equal:

8x – 3 = A(4x – 5) + B(2x + 1).

Take (because that's a little nicer than ) and solve for A:

Then take x = 0 and solve for B:

We get the decomposition

Exercise 6

Decompose into partial fractions.

Answer

We need to find A and B so that

Add the partial fractions and set the resulting numerator equal to the numerator of the original rational function:

11 – 23x = A(2 – 5x) + B(1 – x).

Take x = 1 to solve for A:

11 – 23x = A(2 – 5x) + B(1 – x)

11 – 23(1) = A(2 – 5(1)) + B(1 – 1)

-12 = -3A

4 = A

Then take x = 0 and solve for B:

11 – 23x = A(2 – 5x) + B(1 – x)

11 – 23(0) = (4)(2 – 5(0)) + B(1 – 0)

11 = 8 + B

3 = B

We get the decomposition

Exercise 7

Decompose into partial fractions.

Answer

We need to find A and B so that

Add the partial fractions and set the resulting numerator equal to the numerator of the original function:

3 – 10x = A(1 – x) + B(3x + 4).

Take x = 1 and solve for B:

3 – 10x = A(1 – x) + B(3x + 4)

3 – 10(1) = A(1 – (1)) + B(3(1) + 4)

-7 = 7B

-1 = B

Then take x = 0 and solve for A:

3 – 10x = A(1 – x) + B(3x + 4)

3 – 10(0) = A(1 – (0)) + (-1)(3(0) + 4)

3 = A – 4

7 = A

Instead of writing the decomposition as

we write it as

Exercise 8

Decompose into partial fractions.

Answer

We need to find A and B so that

Add the partial fractions and set the resulting numerator equal to the numerator of the original function:

2x + 4 = A(x + 1) + B(2x + 1).

Take x = -1 and solve for B:

2x + 4 = A(x + 1) + B(2x + 1)

2(-1) + 4 = A(-1 + 1) + B(2(-1) + 1)

2 = -B

-2 = B.

Then take x = 0 and solve for A:

2x + 4 = A(x + 1) + B(2x + 1)

2(0) + 4 = A((0) + 1) + (-2)(2(0) + 1)

4 = A – 2

6 = A

We write the decomposition

Exercise 9

Decompose into partial fractions.

Answer

The denominator factors as

x² + 3x + 2 = (x + 1)(x + 2)

so the partial fraction decomposition will look like

We set the original numerator and the numerator of the sum of the partial fractions equal to each other:

2x + 3 = A(x + 2) + B(x + 1).

We set x = -1 and solve for A:

2x + 3 = A(x + 2) + B(x + 1)

2(-1) + 3 = A(-1 + 2) + B(-1 + 1)

1 = A

We set x = 0 and solve for B:

2x + 3 = A(x + 2) + B(x + 1)

2(0) + 3 = (1)(0 + 2) + B(0 + 1)

3 = 2 + B

1 = B

The decomposition is

Exercise 10

Decompose into partial fractions.

Answer

The denominator factors as

2x² – 5x – 3 = (x – 3)(2x + 1)

so the partial fraction decomposition will look like

Add the partial fractions and set their numerator equal to the original numerator:

8x – 10 = A(2x + 1) + B(x – 3).

Set x = 3 and solve for A:

8x – 10 = A(2x + 1) + B(x – 3)

8(3) – 10 = A(2(3) + 1) + B(3 – 3)

14 = 7A

2 = A

Now set x = 0 and solve for B:

8x – 10 = A(2x + 1) + B(x – 3)

8(0) – 10 = (2)(2(0) + 1) + B(0 – 3)

-10 = 2 – 3B

3B = 12

B = 4

The decomposition is

Exercise 11

Decompose into partial fractions.

Answer

The denominator factors as

-2x² + 7x + 15 = (2x + 3)(5 – x)

or as

-2x² + 7x + 15 = (-2x – 3)(x – 5).

We'll go with

-2x² + 7x + 15 = (2x + 3)(5 – x)

first. Then the decomposition will look like

Add the partial fractions and set the resulting numerator equal to the original numerator:

-18x – 1 = A(5 – x) + B(2x + 3).

Take x = 5 and solve for B:

-18x – 1 = A(5 – x) + B(2x + 3)

-18(5) – 1 = A(5 – 5) + B(2(5) + 3)

-91 = 13B

-7 = B

Take x = 0 and solve for A:

-18x – 1 = A(5 – x) + B(2x + 3)

-18(0) – 1 = A(5 – 0) + (-7)(2(0) + 3)

-1 = 5A – 21

4 = A

The decomposition is

If instead we had started by factoring the denominator as

-2x² + 7x + 15 = (-2x – 3)(x – 5)

then the decomposition would look like

Add the partial fractions and set the resulting numerator equal to the original numerator:

-18x – 1 = A(x – 5) + B(-2x – 3).

Set x = 5 and solve for B:

-18x – 1 = A(x – 5) + B(-2x – 3)

-18(5) – 1 = A(5 – 5) + B(-2(5) – 3)

-91 = -13B

7 = B

Set x = 0 and solve for A:

-18x – 1 = A(x – 5) + B(-2x – 3)

-18(0) – 1 = A(0 – 5) + 7(-2(0) – 3)

-1 = -5A – 21

20 = -5A

-4 = A

Then the decomposition looks like this:

While these two answers may look different, they are in fact the same. If we multiply each term of our first answer by 1, we get

which is the same as our second answer.

Exercise 12

Decompose into partial fractions.

Answer

The denominator factors nicely into

2x² + 14x = 2x(x + 7)

so the decomposition will look like

Add the partial fractions and set the resulting numerator equal to the original numerator:

15x + 21 = A(x + 7) + B(2x).

Set x = 0 and find A:

15x + 21 = A(x + 7) + B(2x)

15(0) + 21 = A(0 + 7) + B(2(0))

21 = 7A

3 = A

Set x = 1 and find B:

15x + 21 = A(x + 7) + B(2x)

15(1) + 21 = (3)(1 + 7) + B(2(1))

36 = 24 + 2B

6 = B

The decomposition is

Exercise 13

Decompose into partial fractions.

Answer

The denominator factors into

3x² + x – 4 = (x – 1)(3x + 4)

so the decomposition into partial fractions will look like

Add the partial fractions and set the resulting numerator equal to the original numerator:

25x + 10 = A(3x + 4) + B(x – 1).

Set x = 1 and solve for A:

25x + 10 = A(3x + 4) + B(x – 1)

25(1) + 10 = A(3(1) + 4) + B(1 – 1)

35 = 7A

5 = A

Set x = 0 and solve for B:

25x + 10 = A(3x + 4) + B(x – 1)

25(0) + 10 = (5)(3(0) + 4) + B(0 – 1)

10 = 20 – B

10 = B

The decomposition is

Exercise 14

Integrate.

Answer

Exercise 15

Integrate.

Answer

This means

Exercise 16

Integrate.

Answer

Exercise 17

Integrate.

Answer

We didn't do this one yet. The denominator factors into

-2x² + 3x – 1 = (-2x + 1)(x – 1)

so the partial fraction decomposition will look like

Add the partial fractions and set the resulting numerator equal to the original numerator:

-3x + 1 = A(x – 1) + B(-2x + 1).

Set x = 1 and solve for B:

-3x + 1 = A(x – 1) + B(-2x + 1)

-3(1) + 1 = A(1 – 1) + B(-2(1) + 1)

-2 = -B

2 = B

Then set x = 0 and solve for A:

-3x + 1 = A(x – 1) + B(-2x + 1)

-3(0) + 1 = A(0 – 1) + (2)(-2(0) + 1)

1 = -A + 2

A = 1

The partial fraction decomposition is

Now we can evaluate the integral:

Exercise 18

Integrate.

Answer

The denominator factors as

6x² + 5x + 1 = (2x + 1)(3x + 1)

so the partial fraction decomposition will look like

Add the partial fractions and set the resulting numerator equal to the original numerator:

-5x – 3 = A(3x + 1) + B(2x + 1).

Set and solve for A:

Set x = 0 and solve for B:

-5x – 3 = A(3x + 1) + B(2x + 1)

-5(0) – 3 = (1)(3(0) + 1) + B(2(0) + 1)

-3 = 1 + B

-4 = B

The partial fraction decomposition is

Now we can integrate.

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