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Prove that $\int_{0}^{\infty}\frac{tan^{-1}ax}{x(1+x^2)}dx\,=\frac{\pi}{2}log(1+a)\,,\,a\geq0\,,a\neq 1 $
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$ \text{Let I(a) ] } = \int_0^{\infty} \frac{ tan^{-1}ax }{x(1+x^2)} dx $

$ \frac{dI}{da} = \int_0^{\infty} \frac{\partial}{\partial a} \frac{ tan^{-1}ax }{x(1+x^2)} dx $

$ = \int_0^{\infty} \frac{ 1}{x(1+x^2)} \frac{x }{(1+x^2)} dx $

$ = \frac{1}{1-a^2} \int_0^{\infty} [ \frac{ 1}{(1+x^2)} - \frac{a^2 }{(1+a^2x^2)} ]$

$ = \frac{1}{1-a^2} \left [ tan^{-1}x - atan^{-1}ax \right ]_0^{\infty} $

$ = \frac{1}{1-a^2} [\frac{\pi}{2} - a \frac {\pi}{2} ] $

$ = \frac{1}{1-a^2} (1-a) \frac {\pi}{2} $

$ = \frac{\pi}{2(1+a) } $

$ \text{I(a)} = \frac{\pi}{2} log(1+ a) + c $

$ a= 0, I(0) = 0 + c \Longrightarrow c = 0 $

$ \text{I(a)} = \frac{\pi}{2} log(1+ a) $

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