<s>[INST] 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?
 [/INST] \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}
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