How to integrate $\int \frac{\arctan x}{x^4} dx$? I have written the integral as $\int x^{-4} \arctan x dx$. Then, by applying by parts, I got $-3\dfrac{\arctan x}{x^3} + 3\int \dfrac{1}{x^3(1 + x^2)} dx$. Now, how can I solve the later integral? Is there any other trick to do this?
 A: Try partial fractions. In particular, we can expand $$\frac{1}{x^{3}(1+x^{2})}=\frac{1}{x^{3}}+\frac{x}{x^{2}+1}-\frac{1}{x}$$
A: 
$$\int\frac{\arctan x}{x^4}dx$$

By parts:
$$=-\frac{\arctan x}{3x^3}+\frac{1}{3}\int\frac{1}{x^3(x^2+1)}dx$$
$u=x^2,\;du=2xdx$
$$=-\frac{\arctan x}{3x^3}+\frac{1}{6}\int\frac{1}{u^2(u+1)}du$$
Partial fractions
$$=-\frac{\arctan x}{3x^3}+\frac{1}{6}\int\bigg(\frac{1}{u^2}+\frac{1}{u+1}-\frac{1}{u}\bigg)du$$
$$=-\frac{1}{6x^3}(x^3\ln(x^2)-x^3\ln(x^2+1)+x+2\arctan (x))+C$$
Which is equivalent for restricted $x$ values to:
$$\boxed{\color{red}{-\frac{1}{6x^3}(2x^3\ln(x)-x^3\ln(x^2+1)+x+2\arctan(x))+C)}}$$
A: \begin{eqnarray}
\int\frac{\arctan x}{x^4}\,dx&=&-\frac{\arctan x}{3x^3}+\frac13\int\frac{1}{x^3(1+x^2)}\,dx\stackrel{x=1/t}{=}-\frac{\arctan x}{3x^3}-\frac13\int\frac{1}{t^{-3}(1+t^{-2})t^2}\,dt\\[10pt]
&=&-\frac{\arctan x}{3x^3}-\frac13\int\frac{t^3}{1+t^2}\,dt=-\frac{\arctan x}{3x^3}-\frac13\int\frac{t(t^2+1)-t}{1+t^2}\,dt\\[10pt]
&=&-\frac{\arctan x}{3x^3}-\frac13\int\left(t-\frac{t}{1+t^2}\right)\,dt=-\frac{\arctan x}{3x^3}-\frac13\left[t^2-\frac12\ln(1+t^2)\right]+C\\[10pt]
&\stackrel{t=1/x}{=}&-\frac{\arctan x}{3x^3}-\frac13\left[\frac{1}{x^2}-\frac12\ln\left(1+\frac{1}{x^2}\right)\right]+C\\[10pt]
&=&-\frac13\left[\frac{\arctan x}{x^3}+\frac{1}{x^2}-\frac12\ln(1+x^2)+\ln|x|\right]+C
\end{eqnarray}
