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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?

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$$\int \theta d\tan\theta = \theta \tan\theta - \int\tan\theta d\theta = \theta \tan\theta + \int\frac{d\cos\theta}{\cos\theta} = \theta \tan\theta + \log\cos\theta + \text{const}. $$ – achille hui Aug 25 '14 at 10:25
What you are missing is $\displaystyle\;\frac{d}{dx}\arctan(\log x) = \frac{1}{\color{red}{x}(1+\log^2 x)}\;$ – achille hui Aug 25 '14 at 10:33

1 Answer 1


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

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

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Thank you, but it's very similar to what I was trying and I end up with the same dead end... Or did I miss something? – user159527 Aug 25 '14 at 10:20
@user159527, The last part of your method should be $$\ln x\frac1{x[(\ln x)^2+1]}$$ Then you can set $\ln x=y$ – lab bhattacharjee Aug 25 '14 at 10:26

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