$I=\int \frac{\sqrt{x^2+1}}{x^4} \log \left(1+\frac{1}{x^2}\right) d x$
Put $x=\tan \theta, d x=\sec ^2 \theta d \theta$
$I=\int \frac{\sqrt{1+\tan ^2 \theta}}{\tan ^4 \theta} \log \left(1+\frac{1}{\tan ^2 \theta}\right) \sec ^2 \theta d \theta$
$=\int \frac{\sec \theta}{\tan ^4 \theta} \cdot\left[\log \left(1+\cot ^2 \theta\right)\right] \sec ^2 \theta d \theta$
$=-2 \int(\log \sin \theta) \cdot \frac{\cos ^2 \theta}{\sin { }^4 \theta} d \theta$
Put $\sin \theta=t, \cos \theta d \theta=d t$
$\therefore \quad I=-2 \int(\log t) \cdot \frac{1}{t^4} d t$
Integrating by parts taking log $t$ as first function
$\begin{aligned}
&I=-2\left[(\log t)-\int t^{-4} d t-\int\left(\frac{d}{d t}(\log t) \int t^{-4} d t\right) d t\right] \\
&=-2\left[-\frac{\log t}{3 t^3}+\frac{1}{3} \int t^{-4} d t\right]=\frac{2}{9}\left[\frac{3 \log t+1}{t^3}\right]+c
\end{aligned}$
Now $\mathrm{t}=\sin \theta$ and $\tan \theta=\mathrm{x}, \mathrm{t}=\sin \theta=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}$
$\begin{aligned}
&\therefore \mathrm{I}=\frac{2\left(1+\mathrm{x}^2\right)^{3 / 2}}{9 \cdot \mathrm{x}^3}\left[3 \log \frac{\mathrm{x}}{\sqrt{\mathrm{x}^2+1}}+1\right]+\mathrm{c} \\
&=-\frac{1}{3}\left(1+\frac{1}{\mathrm{x}^2}\right)^{3 / 2}\left[\log \left(1+\frac{1}{\mathrm{x}^2}\right)-\frac{2}{3}\right]+\mathrm{c}
\end{aligned}$