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