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