Answer
$$\frac{-1}{7}\ln |x|+\frac{1}{8}\ln|x+1|+\frac{1}{56}\ln |x-7|+C$$
Work Step by Step
Given $$\int \frac{d x}{x\left(x^{2}-6 x-7\right)} $$
Since
\begin{align*}
\frac{1}{x\left(x^{2}-6 x-7\right)}&=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{x+1}+\frac{\mathrm{C}}{x-7}\\
&=\frac{A(x+1)(x-7)+Bx(x-7)+Cx(x+1)}{x\left(x^{2}-6 x-7\right)}\\
1&=A(x+1)(x-7)+Bx(x-7)+Cx(x+1)
\end{align*}
Since
\begin{align*}
\text{at}\ x&= 0\ \ \ \ \ \ A=\frac{-1}{7}\\
\text{at}\ x&= -1\ \ \ \ \ \ B=\frac{1}{8}\\
\text{at}\ x&= 7\ \ \ \ \ \ C=\frac{1}{56}
\end{align*}
Then
\begin{align*}
\int \frac{1}{x\left(x^{2}-6 x-7\right)}dx&=\frac{-1}{7}\int\frac{1}{x}dx+\frac{1}{8}\int \frac{1}{x+1}dx+\frac{1}{56}\int \frac{1}{x-7}dx\\
&= \frac{-1}{7}\ln |x|+\frac{1}{8}\ln|x+1|+\frac{1}{56}\ln |x-7|+C
\end{align*}
