TITLE: How can I solve $\int_0^1\frac{\arctan(x^2)}{1+x^2}\,\mathrm dx$?
QUESTION [13 upvotes]: This integral appears very similar to $\int\frac{\arctan x}{1+x^2}\,\mathrm dx$, but this question cannot be solved through the same simple substitution of $u=\arctan x$. WolframAlpha cannot find a symbolic solution to this problem, and this Quora answer
is the only thing I can find that appears to have the exact answer of $\frac14\log^2(1+\sqrt2)$. I am not sure if there is some special trick to be used in solving this, but I have tried everything I know and nothing seems to work.

REPLY [2 votes]: \begin{align}
\int_0^1\frac{\arctan x^2}{1+x^2}dx=&
\int_0^1\frac{1}{1+x^2}\int_0^x\frac{x}{1+x^2y^2}dy\>dx\\
=& \int_0^1\int_y^1 \frac{x}{(1+x^2)(1+x^2y^2)}dx\>dy\\
=& \>\frac12 \int_0^1\frac{\ln\frac{2(1+y^4)}{(1+y^2)^2}}{1-y^2}dy\overset{y\to\frac{1-\sqrt x}{1+\sqrt x}}
=\frac18\int_0^1 \frac{\ln\frac{x^2+6x+1}{x^2+2x+1}}x dx\\
=& \>\frac18\int_0^1 \int_0^{\cosh^{-1}3}\frac{2\sinh t}{x^2+2x\cosh t+1}dt \ dx\\
=&\>\frac18\int_0^{\cosh^{-1}3}t\ dt=\frac1{16}(\cosh^{-1}3)^2
\end{align}