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I would like to evaluate the integral

$$I(a,b) = \int_0^\infty \frac{\tan^{-1} (ax) \tan^{-1} (bx)}{x^{2}}\,dx$$

and I know that

$$\int_0^\infty \frac{\tan^{-1} (rx)}{x(1+x^2)}\,dx = \frac{\pi}{2}\ln (r+1)$$ for $r>0$.

So I start by partial derivative on $a$ and use the last equivalency. Then I get stuck when going back to $I$ since I believe it must also depend on $b$.

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We have that $$ \frac{\partial}{\partial a}\,I(a,b)= \frac{\pi}{2}\log\left(1+\frac{b}{a}\right)$$ hence (by integrating and exploiting $I(a,b)=I(b,a)$): $$\boxed{ \,I(a,b) = \color{red}{\frac{\pi}{2}\left[(a+b)\log(a+b)-a\log a-b\log b\right]}.}$$

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    Could you please explain what you mean by exploiting? – 2017-01-22
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    @ZanHP: by integrating $\frac{\pi}{2}\log(1+b/a)$ with respect to $a$ we get $\frac{\pi}{2}\left[(a+b)\log(a+b)-a\log a-C_b\right]$ but since $I$ is a symmetric function, $C_b=b\log b$. – 2017-01-22
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    The first line isn't obvious. – 2017-01-23