How to find $$ \int \frac{1}{x}\, \tan^{-1} \frac{1}{x} \, dx. $$
I have tried using product rule, by taking $\frac{1}{x}$ as 1st function and $\tan^{-1}\frac{1}{x}$ as second function, but the integration does not end. Any help in this regard is highly appreciated.
$$I=\int\frac{1}{x}\tan ^{-1}\left(\frac{1}{x}\right)\,dx=-\int \frac{1}{t}\tan ^{-1}\left(t\right)\,dt$$ Integration by parts $$I=-\log (t) \tan ^{-1}(t)+\int\frac{\log (t)}{t^2+1}\,dt$$ $$J=\int\frac{\log (t)}{t^2+1}\,dt=\int\frac{\log (t)}{(t+i)(t-i)}\,dt=\frac i 2\int \left(\frac{\log (t)}{t+i}-\frac{\log (t)}{t-i} \right)\,dt$$ Simple change of variable $$\int \frac{\log (t)}{t+a} \,dt=\text{Li}_2\left(-\frac{t}{a}\right)+\log (t) \log \left(1+\frac{t}{a}\right)$$ and then the result given by Wolfram Alpha $$I=\frac{1}{2} i \left(\text{Li}_2\left(\frac{i}{x}\right)-\text{Li}_2\left(-\frac {i}{x}\right)\right)$$
Edit
If we use the series expansion of $\tan ^{-1}\left(t\right)$, we have $$I=\int \sum_{n=0}^\infty (-1)^{n+1} \frac{t^{2 n}}{2 n+1}\,dt= \sum_{n=0}^\infty (-1)^{n+1} \frac{t^{2 n+1}}{(2 n+1)^2}= -\frac{t}{4} \,\Phi \left(-t^2,2,\frac{1}{2}\right)$$ where appears the Lerch transcendent function