<s>[INST] Integral with partial fractions I'm trying to calculate the following integral:

$$I=\int \frac{\arctan(x)}{x^4}dx$$

My steps so far are:
Per partes:
$$\frac{-\arctan(x)}{3x^3}+\int{\frac{1}{1+x^2} \frac{1}{3x^3}}dx
=\frac{-\arctan(x)}{3x^3}+\frac{1}{3}\int{\frac{1}{1+x^2} \frac{1}{x^3}}dx$$
and now I want to do partial fractions. However, with this integral, I fail to do partial fractions. Could you help me?
Thanks
 [/INST] Without partial fractions
$$
\begin{aligned}
I=\int \frac{\arctan x}{x^4} d x & =-\frac{1}{3} \int \arctan x d\left(\frac{1}{x^3}\right) \\
& \stackrel{IBP}{=}  -\frac{\arctan x}{3 x^3}+\frac{1}{3} \underbrace{\int \frac{1}{x^3\left(1+x^2\right)}}_J dx
\end{aligned}
$$
Letting $y=\frac{1}{x}  $ yields
$$
\begin{aligned}
J & =\int \frac{1}{\frac{1}{y^3}\left(1+\frac{1}{y^2}\right)}\left(-\frac{d y}{y^2}\right) \\
& =-\int \frac{y^3}{y^2+1} d y \\
& =-\int \frac{y\left(y^2+1\right)-y}{y^2+1} d y \\
& =-\frac{y^2}{2}+\frac{1}{2} \ln \left(y^2+1\right)+C
\end{aligned}
$$
Hence $$
I=\frac{1}{6}\left[-\frac{2 \arctan x}{x^3}-\frac{1}{x^2}+\ln \left(1+x^2\right)-2 \ln |x|\right]+C
$$
</s>