# How do you use partial fraction decomposition to decompose the fraction to integrate (6x)/((x-4) (x+4))?

Mar 28, 2018

The integral is $3 \ln | {x}^{2} - 16 | + C$

#### Explanation:

We have:

$\frac{A}{x - 4} + \frac{B}{x + 4} = \frac{6 x}{\left(x - 4\right) \left(x + 4\right)}$

$A \left(x + 4\right) + B \left(x - 4\right) = 6 x$

$A x + B x + 4 A - 4 B = 6 x$

Thus $\left\{\begin{matrix}A + B = 6 \\ 4 A - 4 B = 0\end{matrix}\right.$

Solving we get

$2 A = 6$

$A = 3$

$B = 3$

Therefore the integral becomes $\int \frac{3}{x - 4} + \frac{3}{x + 4} \mathrm{dx}$. This is readily integrated as $3 \ln | x - 4 | + 3 \ln | x + 4 | = 3 \ln | {x}^{2} - 16 | + C$#

Hopefully this helps!