Answer
$$\frac{1}{25} \ln |x|-\frac{1}{50} \ln \left|x^{2}+25\right|+C$$
Work Step by Step
Given $$\int \frac{d x}{x\left(x^{2}+25\right)}$$
Since
\begin{align*}
\frac{1}{(x )\left(x^{2}+25\right)}&=\frac{A}{x }+\frac{Bx+C}{x^{2}+25}\\
&=\frac{A(x^{2}+25)+B(x)}{(x)\left(x^{2}+25\right)}\\
1&=A(x^{2}+25)+B(x)
\end{align*}
\begin{align*}
\text{at } x&=0 \ \ \ \ \to A=\frac{1}{25} \\
\text{at } x&=1\ \ \ \ \to B+C=\frac{-1}{25}\\
\text{at } x&= -1\ \ \ \ \to B -C=\frac{-1}{25}
\end{align*}
Hence $B=\frac{-1}{25}$ and
\begin{aligned}
\int \frac{1}{x\left(x^{2}+25\right)} d x &=\int \frac{\frac{1}{25}}{x} d x-\int \frac{ \frac{1}{25} x}{x^{2}+25} d x \\
&=\frac{1}{25} \ln |x|-\frac{1}{50} \ln \left|x^{2}+25\right|+C
\end{aligned}
