<s>[INST] Solving the Integral using $\ln|u|$ Can Someone help me solve this
$$
\int\frac{19\tan^{-1}x}{x^{2}}\,dx
$$
We have been told to use $\ln|u|$ and $C$.
Thanks!
 [/INST] Using integration by parts we see:
$$ \int \frac{ \arctan(x) }{x^2} dx = -\frac{\arctan(x) }{x} + \int \frac{dx}{ (x^2+1) x} $$
Now if we use a partial fraction decomposition we obtain:
$$\int \frac{dx}{ (x^2+1) x} = \int \left (  \frac{1}{x} - \frac{x}{x^2 +1} \right ) dx $$
From here it should be clear that
$$\int \frac{ \arctan(x) }{x^2} dx = \ln (x) - \frac{1}{2} \ln (x^2 +1) - \frac{\arctan(x)}{x} +C $$
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