Answer
$$-\frac{1}{25 x}-\frac{1}{125} \tan ^{-1}\left(\frac{x}{5}\right)+C$$
Work Step by Step
Given $$\int \frac{d x}{x^2\left(x^{2}+25\right)}$$
Since
\begin{align*}
\frac{1}{(x^2 )\left(x^{2}+25\right)}&=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C x+D}{x^{2}+25}\\
&=\frac{A x^{3}+25 A x+B x^{2}+25 B+C x^{3}+D x^{2}}{(x^2 )\left(x^{2}+25\right)} \\
1&=A x^{3}+25 A x+B x^{2}+25 B+C x^{3}+D x^{2}
\end{align*}
Then by comparing the coefficients, we get
$$A=C=0, \ \ \ B=\frac{1}{25},\ \ \ C=\frac{-1}{25}$$
Hence $B=\frac{-1}{25}$ and
\begin{aligned}
\int \frac{1}{x^2\left(x^{2}+25\right)} d x &=\frac{1}{25} \int \frac{1}{x^2} d x-\frac{1}{25}\int \frac{ 1}{x^{2}+25} d x \\
&= -\frac{1}{25 x}-\frac{1}{125} \tan ^{-1}\left(\frac{x}{5}\right)+C
\end{aligned}
