Answer
$$2 \ln |x-2|+4 \ln |x+2|+\frac{1}{x+2}+C$$
Work Step by Step
Given $$\int \frac{6 x^{2}+7 x-6}{\left(x^{2}-4\right)(x+2)} d x$$
Since \begin{aligned}
\frac{6 x^{2}+7 x-6}{\left(x^{2}-4\right)(x+2)} &=\frac{6 x^{2}+7 x-6}{(x-2)(x+2)(x+2)} \\
&=\frac{6 x^{2}+7 x-6}{(x-2)(x+2)^{2}} \\
&=\frac{A}{(x-2)}+\frac{B}{(x+2)}+\frac{C}{(x+2)^{2}} \\
6 x^{2}+7 x-6 &=A(x+2)^{2}+B(x-2)(x+2)+C(x-2)
\end{aligned}
Then\begin{align*}
\text{at } x&=2 \ \ \ \ \to A=2 \\
\text{at } x&= -2\ \ \ \ \to C=-1\\
\text{at } x&= 0\ \ \ \ \to B=4
\end{align*}
Hence
\begin{aligned}
\int \frac{6 x^{2}+7 x-6}{\left(x^{2}-4\right)(x+2)} d x &=\int \frac{2}{x-2} d x+\int \frac{4}{x+2} d x-\int \frac{1}{(x+2)^{2}} d x \\
&=\int \frac{2}{x-2} d x+\int \frac{4}{x+2} d x-\int(x+2)^{-2} d x\\
&=2 \ln |x-2|+4 \ln |x+2|+\frac{1}{x+2}+C
\end{aligned}
