Answer
$$3 \ln |x-5|+\ln |x+3|+C$$
Work Step by Step
Given $$\int \frac{4 x+4}{(x-5)(x+3)} d x$$
Since
\begin{aligned}
\frac{4 x+4}{(x-5)(x+3)} &=\frac{A}{(x-5)}+\frac{B}{(x+3)} \\
&= \frac{A(x+3)+B(x-5)}{(x-5)(x+3)} \\
4 x+4 &=A(x+3)+B(x-5)
\end{aligned}
Then
\begin{align*}
\text{at }\ x&= 5\ \ \ \ \ \ \ \ A= 3\\
\text{at }\ x&= -3\ \ \ \ \ \ \ \ B= 1
\end{align*}
Hence \begin{aligned}
\int \frac{4 x+4}{(x-5)(x+3)} d x &=\int \frac{3}{(x-5)} d x+\int \frac{1}{(x+3)} d x \\
&=3 \ln |x-5|+\ln |x+3|+C
\end{aligned}
