Integration using partial fractions

How do you integrate an expression when there is an algebraic expression in the numerator and denominator of a fraction? Integrating using partial fractions helps you to solve this problem. Use this resource to learn how to integrate using partial fractions.

Sometimes a complex function may be integrated by breaking it up into partial fractions. The idea is that each partial fraction is an easier integral than the original.

For example:
$\int \frac{1}{x^{2} + 5 x + 6} d x = \int \frac{1}{x + 2} d x - \int \frac{1}{x + 3} d x$

The integrals on the right are much simpler than that on the left.

Before you dive into this resource, make sure you are confident with finding partial fractions.

Example 1 – integrating using partial fractions

Find the integral of $\frac{x - 5}{x^{2} + 2 x - 3}$ with respect to $x$ .

First, find the partial fractions. Since $x^{2} + 2 x - 3 = (x + 3) (x - 1)$ , we are dealing with the case where the denominator has two distinct linear factors. Therefore, we factorise as follows:
$\begin{aligned} \frac{x - 5}{x^{2} + 2 x - 3} & = \frac{x - 5}{(x + 3) (x - 1)} \\ = \frac{A}{x + 3} + \frac{B}{x - 1} \end{aligned}$

Now, we have to determine the constants $A$ and $B$ . First, add the partial fractions by finding a common denominator. In this case, it is $(x + 3) (x - 1)$ .
$\begin{aligned} \frac{A}{x + 3} + \frac{B}{x - 1} & = \frac{A}{(x + 3)} \times \frac{(x - 1)}{(x - 1)} + \frac{B}{(x - 1)} \times \frac{(x + 3)}{(x + 3)} \\ = \frac{A (x - 1) + B (x + 3)}{(x + 3) (x - 1)} \\ = \frac{A x - A + B x + 3 B}{(x + 3) (x - 1)} \\ = \frac{A x + B x - A + 3 B}{x^{2} + 2 x - 3} \\ = \frac{(A + B) x - A + 3 B}{x^{2} + 2 x - 3} \end{aligned}$

Equating the numerators:
$x - 5 = (A + B) x - A + 3 B$

where $A + B$ is the $x$ term and $- A + 3 B$ is the constant. On the left, the coefficient of $x$ is $1$ and on the right, the coefficient is $A + B$ . Similarly, the constant term on the left is $- 5$ and on the right, it is $- A + 3 B$ .

This gives us the following simultaneous equations:
$\begin{aligned} A + B & = 1 \\ - A + 3 B & = - 5 \end{aligned}$

These equations are easily solved by elimination. Adding the equations, we get:
$\begin{aligned} 4 B & = - 4 \\ B & = - 1 \end{aligned}$

Substituting for $B$ the first equation gives:
$\begin{aligned} A - 1 & = 1 \\ A & = 2 \end{aligned}$

Finally, we have:
$\frac{x - 5}{x^{2} + 2 x - 3} = \frac{2}{x + 3} - \frac{1}{x - 1}$

We can now evaluate the integral:
$\begin{aligned} \int \frac{x - 5}{x^{2} + 2 x - 3} d x & = \int (\frac{2}{x + 3} - \frac{1}{x - 1}) d x \\ = \int \frac{2}{x + 3} d x - \int \frac{1}{x - 1} d x \\ = 2 \ln | x + 3 | - \ln | x - 1 | + c, c \in R \\ = \ln (x + 3)^{2} - \ln | x - 1 | + c \\ = \ln \frac{(x + 3)^{2}}{| x - 1 |} + c \end{aligned}$

Find the integral of $\frac{x^{2} - 3 x + 16}{x^{3} - 5 x^{2} + x - 5}$ with respect to $x$ .

We, first factorise the denominator.
$\begin{aligned} x^{3} - 5 x^{2} + x - 5 & = x^{2} (x - 5) + x - 5 \\ = (x^{2} + 1) (x - 5) \end{aligned}$

In this case, we have a quadratic factor and a linear factor. Therefore, we factorise as follows:
$\begin{aligned} \frac{x^{2} - 3 x + 16}{x^{3} - 5 x^{2} + x - 5} & = \frac{A x + B}{x^{2} + 1} + \frac{C}{x + 5} \\ = \frac{(A x + B) (x - 5)}{(x^{2} + 1) (x - 5)} + \frac{C (x^{2} + 1)}{(x^{2} + 1) (x - 5)} \\ = \frac{A x^{2} - 5 A x + B x - 5 B + C x^{2} + C}{(x^{2} + 1) (x + 5)} \\ = (A + C) x^{2} + (- 5 A + B) x - 5 B + C \end{aligned}$

Equating coefficients of $x^{2}$ , $x$ and the constant gives the three equations:
$\begin{aligned} A + C & = 1 \\ - 5 A + B & = - 3 \\ - 5 B + C & = 16. \end{aligned}$

To solve them simultaneously, we can first multiply the first equation by $5$ .
$5 A + 5 C = 5$

We can add the second equation to this new equation to get: $\begin{aligned} B + 5 C & = - 3 + 5 \\ = 2 \\ 5 B + 25 C & = 10 \end{aligned}$

Adding the same equation to the third equation gives:
$\begin{aligned} 26 C & = 26 \\ C & = 1 \end{aligned}$

Substituting $C = 1$ into the first equation gives:
$\begin{aligned} A + 1 & = 1 \\ A & = 0 \end{aligned}$

Finally, substituting $A = 0$ into the second equation gives:
$B = - 3$

Therefore, we have:
$\frac{x^{2} - 3 x + 16}{x^{3} - 5 x^{2} + x - 5} = \frac{- 3}{x^{2} + 1} + \frac{1}{x - 5}$

We can now find the integral:
$\begin{aligned} \int (\frac{x^{2} - 3 x + 16}{x^{3} - 5 x^{2} + x - 5}) d x & = \int (\frac{- 3}{x^{2} + 1} + \frac{1}{x - 5}) d x \\ = \int \frac{1}{x - 5} d x - 3 \int \frac{1}{x^{2} + 1} d x \\ = \ln | x - 5 | - 3 \tan^{- 1} (x) + c, c \in R \end{aligned}$

Exercise – integrating using partial fractions

Perform the following integrations.

$\int \frac{4 x + 9}{x^{2} + x - 12} d x$
$\int \frac{21 - 8 x}{x^{2} - x - 6} d x$
$\int \frac{5 x^{2} - 5 x + 2}{(x + 1) (x - 1)^{2}} d x$

$\ln ({| 3 - x |}^{3} | x + 4 |) + c$
$- \frac{3}{5} \ln | 3 - x | - \frac{37}{5} \ln | x + 2 | + c$
$\frac{1}{1 - x} + \ln ((1 - x)^{2} {| x + 1 |}^{3} + c)$

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Integration using partial fractions

Example 1 – integrating using partial fractions

Exercise – integrating using partial fractions

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Integration using partial fractions

Example 1 – integrating using partial fractions

Example 2

Exercise – integrating using partial fractions

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