# How do you integrate int (2x^3 - 4x^2 -15x + 5) / (x^2 -2x - 8) using partial fractions?

Nov 19, 2016

$= {x}^{2} + \frac{3}{2} \ln | x - 4 | - \frac{1}{2} \ln | x + 2 | + C$

#### Explanation:

First divide the numerator by the denominator using long division. So, our new integral is $\int \left(2 x + \frac{x + 5}{{x}^{2} - 2 x - 8}\right) \mathrm{dx}$

We now can write $\frac{x + 5}{{x}^{2} - 2 x - 8}$ in partial fractions.

First of all, ${x}^{2} - 2 x - 8$ can be factored as $\left(x - 4\right) \left(x + 2\right)$.

$\frac{A}{x - 4} + \frac{B}{x + 2} = \frac{x + 5}{\left(x - 4\right) \left(x + 2\right)}$

$A \left(x + 2\right) + B \left(x - 4\right) = x + 5$

$A x + 2 A + B x - 4 B = x + 5$

$\left(A + B\right) x + \left(2 A - 4 B\right) = x + 5$

We can now write a system of equations.

$\left\{\begin{matrix}A + B = 1 \\ 2 A - 4 B = 5\end{matrix}\right.$

Solving:

$B = 1 - A$

$2 A - 4 \left(1 - A\right) = 5$

$2 A - 4 + 4 A = 5$

$6 A = 9$

$A = \frac{3}{2}$

$B = - \frac{1}{2}$

Hence, the partial fraction decomposition of $\frac{x + 5}{{x}^{2} - 2 x - 8}$ is $\frac{3}{2 \left(x - 4\right)} - \frac{1}{2 \left(x + 2\right)}$

The integral now becomes:

$\int \left(2 x + \frac{3}{2 \left(x - 4\right)} - \frac{1}{2 \left(x + 2\right)}\right) \mathrm{dx}$

Which can be readily integrated as;

${x}^{2} + \frac{3}{2} \ln | x - 4 | - \frac{1}{2} \ln | x + 2 | + C$ using the rule $\int \left(\frac{1}{u}\right) \mathrm{du} = \ln | u | + C$ and $\int \left({x}^{n}\right) \mathrm{dx} = \frac{{x}^{n + 1}}{n + 1}$. You can check the result by differentiating.

Hopefully this helps!