Answer
$$5 \ln |x-1|+\ln |x+1|+C$$
Work Step by Step
Given $$\int \frac{(6 x+4) d x}{x^{2}-1}$$
Since
\begin{aligned}
\frac{6 x+4}{x^{2}-1}&=\frac{A}{(x-1)}+\frac{B}{(x+1)}\\
6 x+4&=A(x+1)+B(x-1)
\end{aligned}
Then
\begin{align*}
\text{at }\ x &= 1\ \ \ \ \ \ \ \ \ \ A=5\\
\text{at }\ x &=-1\ \ \ \ \ \ \ \ \ \ B=1
\end{align*}
Hence
\begin{aligned}
\int \frac{6 x+4}{x^{2}-1} d x &=5 \int \frac{1}{(x-1)} d x+\int \frac{1}{(x+1)} d x \\
&=5 \ln |x-1|+\ln |x+1|+C
\end{aligned}
