# Integrate solution: \int(\frac{\tan^{-1}x}{x-\tan^{-1}x})^2dx

Integrate solution:
$\int {\left(\frac{{\mathrm{tan}}^{-1}x}{x-{\mathrm{tan}}^{-1}x}\right)}^{2}dx$
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louran20z47
With $x=\mathrm{tan}\left(u\right),dx=\frac{1}{{\mathrm{cos}}^{2}\left(u\right)}du$, thus
$\int {\left(\frac{{\mathrm{tan}}^{-1}x}{x-{\mathrm{tan}}^{-1}x}\right)}^{2}dx=\int {\left(\frac{u}{\mathrm{tan}\left(u\right)-u}\cdot \frac{1}{\mathrm{cos}\left(u\right)}\right)}^{2}du$
Now, observe that
$d\left(\frac{1}{\mathrm{sin}\left(u\right)-u\mathrm{cos}\left(u\right)}\right)=\frac{-u\mathrm{sin}\left(u\right)du}{{\left(\mathrm{sin}\left(u\right)-u\mathrm{cos}\left(u\right)\right)}^{2}}$
This allows to integrate by parts:
$=-\frac{1}{x}\left(1+\frac{\left(1+{x}^{2}\right)\mathrm{arctan}\left(x\right)}{x-\mathrm{arctan}\left(x\right)}\right)+C=-\frac{1+x\mathrm{arctan}\left(x\right)}{x-\mathrm{arctan}\left(x\right)}+C$