Resembling Resemblence -2

It looks really hard and forces you to think hard but a really big heartbreak comes when you notice the actual solution.

$\displaystyle I(a) = \int_{0}^{\infty}{\frac{log\left(1+{x}^{a}\right)}{\left(1+{x}^{2}\right)log x} \ dx}$

Now $x \rightarrow \frac{1}{x}$ ,

$\displaystyle I(a) = \int_{0}^{\infty}{\frac{log\left(1+{x}^{a}\right)}{\left(1+{x}^{2}\right)log\left(\frac{1}{x}\right)} \ dx} + \int_{0}^{\infty}{\frac{log({x}^{a})}{(1+{x}^{2}) log x} \ dx}$

$\displaystyle I(a) = - I(a) + a \int_{0}^{\infty}{\frac{1}{1+{x}^{2}} \ dx}$

$\displaystyle I(a) = \frac{a}{2}\left({tan}^{-1}(\infty) - {tan}^{-1}(0)\right)$

$\displaystyle I(a) = \boxed{\frac{a \ \pi}{4}}$