# How do you integrate int x / ((x-5)(x+3))dx using partial fractions?

Feb 1, 2016

I found: $\frac{1}{8} \left[5 \ln | x - 5 | + 3 \ln | x + 3 |\right] + c$

#### Explanation:

The argument can be written as:
$\frac{x}{\left(x - 5\right) \left(x + 3\right)} = \frac{1}{8} \left[\frac{5}{x - 5} + \frac{3}{x + 3}\right]$
The integral becomes:
$\frac{1}{8} \int \left[\frac{5}{x - 5} + \frac{3}{x + 3}\right] \mathrm{dx} =$
That can be written as:
=1/8[int5/(x-5)dx+int3/(x+3)dx=1/8[5ln|x-5|+3ln|x+3|]+c