Evaluating the indefinite integral $\int\log\left(\tan^{-1}(x)\right)\;\mathrm dx$ I have tried to solve this and have solved it. But I can't.
$$\int\log\left(\tan^{-1}(x)\right)\;\mathrm dx $$
I got this answer $$ \log\left(\tan^{-1}(x)\right)\cdot x. $$ But its wrong. Please help me someone.
 A: Using the substitution $\tan(u)=x$, we get
$$
\begin{align}
\int\log(\tan^{-1}(x))\,\mathrm{d}x
&=\int\log(u)\,\mathrm{d}\tan(u)\\
&=\log(u)\tan(u)-\int\frac{\tan(u)}{u}\mathrm{d}u
\end{align}
$$
However, I do not believe that $\frac{\tan(u)}{u}$ has an elementary anti-derivative.
A: $ \int \log(\tan^{-1}(x))\,\mathrm{d}x$ Let $\tan^{-1}(x) = t \Rightarrow \frac{1}{1+x^2}\mathrm{d}x = \mathrm{d}t $ .....(i)
Also x =\tan(t) .......(ii)  
Therefore ( by using (ii)):
$$
\begin{align}
\int\log(\tan^{-1}(x))\,\mathrm{d}x
&= \int\log(t) . (1+\tan^2(t))\,\mathrm{d}t\\
&=\int (\log(t) + \log(t) \tan^2(t))dt\\
&= \int( \log(t) + \log(t) \sec^2(t) - \log(t) )\,\mathrm{d}t\\
&= \int \log(t) \sec^2(t)\,\mathrm{d}t 
\end{align}
$$
Which by using integration by parts can easily be solved  and using (ILATE ) which means take Inverse function, L-logarithmic function, A-algebric functin, T- Trigonometric function, E - exponential function . 
Which can be written as :
$$
\log(t) \int \sec^2(t)\,\mathrm{d}t -\int \frac{1}{t}\tan(t)\,\mathrm{d}t
= \log(t)\tan(t) -\int \frac{1}{t}\tan(t)\,\mathrm{d}t
$$   from here you can easily solve this.... 
