Show $\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx= \int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$ I accidentally found out that the two integrals below
$$I_1=\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx,\>\>\>\>\>\>\>I_2=\int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$$
are equal in value. In fact, they can be evaluated explicitly. For example, the first one can be carried out via double integration,  as sketched below.
\begin{align}
I_1&=\int_0^\infty \left(\int_0^1 \frac{x^2}{1+y^2x^4}dy\right)\frac{1}{1+x^2}dx\\
&= \frac\pi2\int_0^1 \left( \sqrt{\frac y2}+ \frac1{\sqrt{2y} }-1\right)\frac{1}{1+y^2}dy=\frac{\pi^2}8
\end{align}
Similarly, the second one yields $I_2=\frac{\pi^2}8$ as well.
The evaluations are a bit involved, though, and it seems an overreach to prove their equality this way, if only the following needs to be shown
$$\int_0^\infty \frac{\tan^{-1}x^2-\tan^{-1}x^{1/2} }{1+x^2}dx=0$$
The question, then, is whether there is a shortcut to show that the above integral vanishes.
 A: 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*}
A: 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 $k\in\Bbb R$.
A: Let
$$J=\int_0^\infty \frac{\tan^{-1}x^2-\tan^{-1}x^{1/2} }{1+x^2}dx.$$
Using
$$ \arctan x+\arctan\frac1x=\frac{\pi}2$$
and under $x\to \frac1x$, one has
\begin{eqnarray}
J&=&\int_0^\infty \frac{\tan^{-1}x^2-\tan^{-1}x^{1/2} }{1+x^2}dx\\
&=&\int_0^\infty \frac{\tan^{-1}x^{-2}-\tan^{-1}x^{-1/2} }{1+x^2}dx\\
&=&\int_0^\infty \frac{\tan^{-1}x^{1/2}-\tan^{-1}x^{2} }{1+x^2}dx\\
&=&-J.
\end{eqnarray}
So $J=0$.
