Show $\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx= \int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$

By the known formula for the difference of two inverse tangents, your last integral is $$ \int_0^{ + \infty } {\frac{{\tan ^{ - 1} \left( {\frac{{x^2 - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}dx} = \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{x^2 - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}dx} + \int_1^{ + \infty } {\frac{{\tan ^{ - 1} \left( {\frac{{x^2 - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}dx} . $$ Taking $y=1/x$ in the last integral gives \begin{align*} & \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{x^2 - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}dx} + \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{y^{1/2} - y^2 }}{{1 + y^{5/2} }}} \right)}}{{1 + y^2 }}dy} \\ & = \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{x^2 - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}dx} - \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{y^2 - y^{1/2} }}{{1 + y^{5/2} }}} \right)}}{{1 + y^2 }}dy} = 0. \end{align*}

Since $\int_0^\infty\frac{f(x)dx}{1+x^2}=\int_0^1\frac{f(x)+f(1/x)}{1+x^2}dx$, $\int_0^\infty\tfrac{\arctan x^kdx}{1+x^2}=\tfrac{\pi}{2}\int_0^1\tfrac{dx}{1+x^2}=\tfrac{\pi^2}{8}$ for all $k\in\Bbb R$.