# Integral by Partial Fractions

## Key Questions

• Here's an example of decomposing a rational function (followed by integration, which is one of the main reasons for partial fraction decomposition):

$\frac{1}{x} - \frac{1}{x + 1} - \frac{1}{x + 1} ^ 2$

#### Explanation:

$\frac{1}{{x}^{3} + 2 {x}^{2} + x} = \frac{1}{x \cdot \left({x}^{2} + 2 x + 1\right)} = \frac{1}{x \cdot {\left(x + 1\right)}^{2}}$

Hence,

$\frac{1}{{x}^{3} + 2 {x}^{2} + x} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x + 1} ^ 2$

After expanding denominator,

$A \cdot {\left(x + 1\right)}^{2} + B \cdot \left({x}^{2} + x\right) + C x = 1$

Set $x = - 1$, $- C = 1$, so $C = - 1$

Set $x = 0$, $A = 1$

Set $x = 1$, $- 4 A + 2 B + C = 1$, so $B = - 1$

Thus,

$\frac{1}{{x}^{3} + 2 {x}^{2} + x} = \frac{1}{x} - \frac{1}{x + 1} - \frac{1}{x + 1} ^ 2$

• This integral can be solved by using the Partial Fractions approach, giving an answer of

$2 \ln \left(x + 5\right) - \ln \left(x - 2\right) + C$

Process:

The partial fractions approach is useful for integrals which have a denominator that can be factored but not able to be solved by other methods, such as Substitution. This equation already has its denominator factored, but note that if we were instead given the multiplied form:

$\int \frac{x - 9}{{x}^{2} + 3 x - 10}$,

we would need to factor the denominator to continue. We can now turn this function into its partial fraction equivalent:

$\frac{A}{x + 5} + \frac{B}{x - 2}$ = $\frac{x - 9}{\left(x + 5\right) \left(x - 2\right)}$

Multiplying by the common denominator: $\left(x + 5\right) \left(x - 2\right)$, we have:

$A \left(x - 2\right) + B \left(x + 5\right) = x - 9$

Now we can choose any value of $x$ to plug in on both sides, so the best solution is to choose values which will cancel out one of the terms on the left side. In this case, they will be $2$ and $- 5$.

Plugging in $- 5$, we have:

$A \left(- 5 - 2\right) + B \left(- 5 + 5\right) = - 5 - 9$
$A \left(- 7\right) = - 14$
$A = 2$

With $2$, we have:

$A \left(2 - 2\right) + B \left(2 + 5\right) = 2 - 9$
$B \left(7\right) = - 7$
$B = - 1$

Using these values in our original partial fractions representation from above, we have:

$\frac{A}{x + 5} + \frac{B}{x - 2}$ = $\frac{2}{x + 5} - \frac{1}{x - 2}$

Now you can integrate these terms separately using substitution. Both will end up being ${u}^{- 1}$, resulting in $\ln u + C$. Now you can arrive at your answer:

$2 \ln \left(x + 5\right) - \ln \left(x - 2\right) + C$

• I would try Integration by Partial Fractions when an integrand is a rational function with its denominator can be factored out into smaller factors.