# How do you find int ( 3x-1)/(x^2+2x-8) dx using partial fractions?

Jun 6, 2017

$\frac{13 \ln \left\mid x + 4 \right\mid + 5 \ln \left\mid x - 2 \right\mid}{6}$

#### Explanation:

$\int \frac{3 x - 1}{{x}^{2} + 2 x - 8} \mathrm{dx} = \int \frac{3 x - 1}{\left(x + 4\right) \left(x - 2\right)} \mathrm{dx}$

To decompose the function into partial fractions, there are $A$ and $B$ such that:

$\frac{A}{x + 4} + \frac{B}{x - 2} = \frac{3 x - 1}{\left(x + 4\right) \left(x - 2\right)}$

$A \left(x - 2\right) + B \left(x + 4\right) = 3 x - 1$
$\left(A + B\right) x + \left(4 B - 2 A\right) = 3 x - 1$

Now solve for $A$ and $B$.
$A + B = 3$
$4 B - 2 A = - 1$
$\to A = \frac{13}{6} , B = \frac{5}{6}$

Integrate:
$\int \frac{\frac{13}{6}}{x + 4} \mathrm{dx} + \int \frac{\frac{5}{6}}{x - 2} \mathrm{dx}$

$\frac{13}{6} \int \frac{1}{x + 4} \mathrm{dx} + \frac{5}{6} \int \frac{1}{x - 2} \mathrm{dx}$

Set $u = x + 4$ in the first and $v = x - 2$ in the second so that both have derivative of $1$.

$\frac{13}{6} \int \frac{1}{u} \mathrm{du} + \frac{5}{6} \int \frac{1}{v} \mathrm{dv}$

$\frac{13}{6} \ln u + \frac{5}{6} \ln v$

$\frac{13}{6} \ln \left\mid x + 4 \right\mid + \frac{5}{6} \ln \left\mid x - 2 \right\mid$

$\frac{13 \ln \left\mid x + 4 \right\mid + 5 \ln \left\mid x - 2 \right\mid}{6}$