How do I proceed with this integral?

Question

I have the following integral: $$\int \frac{\tan^{-1}(\ln (x))}{x}dx.$$ Trying to solve it by integration by parts (with $u=\ln (x)$ and $v=\tan^{-1} (\ln (x))$, I have seemingly come to a dead end: $$\ln (x)\cdot \tan^{-1}(\ln (x))-\int \frac{\ln (x)}{\ln^2(x+1)}dx.$$ How do I proceed? Or did I get it all wrong?

Answer 1

HINT:

Setting $\ln x=y$ $$\int\frac{\arctan(\ln x)}x dx=\int\arctan y\ dy$$

Now integrate by Parts, $$\int\arctan y\ dy=\arctan y\int dy-\int\left(\frac{d(\arctan y)}{dy}\cdot\int dy\right)dy$$