# How do you integrate (x - 16) / (x^2 + x - 2) using partial fractions?

Oct 9, 2016

$\int \left(\frac{x - 16}{{x}^{2} + x - 2}\right) \mathrm{dx} = 6 \ln | x + 2 | - 5 \ln | x - 1 | + c$

#### Explanation:

We first expand the given expression into partial fractions:

$\frac{x - 16}{{x}^{2} + x - 2} \equiv \frac{x - 16}{\left(x + 2\right) \left(x - 1\right)}$
$\frac{x - 16}{{x}^{2} + x - 2} \equiv \frac{A}{x + 2} + \frac{B}{x - 1}$
$\frac{x - 16}{{x}^{2} + x - 2} \equiv \frac{A \left(x - 1\right) + B \left(x + 2\right)}{\left(x + 2\right) \left(x - 1\right)}$

And so. $\left(x - 16\right) \equiv A \left(x - 1\right) + B \left(x + 2\right)$

Put $x = 1 \implies 1 - 16 = 0 + B \left(3\right)$
$\therefore 3 B = - 15 \implies B = - 5$

Put $x = - 2 \implies - 2 - 16 = A \left(- 2 - 1\right) + 0$
$\therefore - 3 A = - 18 \implies A = 6$

So the partial fraction decomposition is:
$\frac{x - 16}{{x}^{2} + x - 2} \equiv \frac{6}{x + 2} - \frac{5}{x - 1}$

We now want to integrate; so
$\int \left(\frac{x - 16}{{x}^{2} + x - 2}\right) \mathrm{dx} = \int \left(\frac{6}{x + 2} - \frac{5}{x - 1}\right) \mathrm{dx}$
$\int \left(\frac{x - 16}{{x}^{2} + x - 2}\right) \mathrm{dx} = 6 \int \left(\frac{1}{x + 2}\right) \mathrm{dx} - 5 \int \left(\frac{1}{x - 1}\right) \mathrm{dx}$
$\int \left(\frac{x - 16}{{x}^{2} + x - 2}\right) \mathrm{dx} = 6 \ln | x + 2 | - 5 \ln | x - 1 | + c$