A definite integral containing arctan (x) I am a university student and today I face to a definite integral containing arctan(x). I can't solve it. As follow:
$$\int_{-1}^1 \frac{x^2 \arctan^2 (x) + \tan(x)}{1+x^2} \, dx.$$
Thanks for any help :)
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Following $\large\tt @Lucian$ comment:
\begin{align}
&2\int_{0}^{\pi/4}x^{2}\tan^{2}\pars{x}\,\dd x
=
-2\int_{0}^{\pi/4}x^{2}\,\dd x + 2\int_{0}^{\pi/4}x^{2}\sec^{2}\pars{x}\,\dd x
\\[3mm]&=-\,{\pi^{3} \over 96} + 2\pars{\pi \over 4}^{2}\tan\pars{\pi \over 4}
-4\int_{0}^{\pi/4}x\tan\pars{x}\,\dd x
\\[3mm]&=-\,{\pi^{3} \over 96} + {\pi^{2} \over 8}
-4\,{\pi \over 4}\ln\pars{\cos\pars{\pi \over 4}}
+4\int_{0}^{\pi/4}\ln\pars{\cos\pars{x}}\,\dd x
\\[3mm]&=-\,{\pi^{3} \over 96} + {\pi^{2} \over 8}
+ {\pi \over 2}\,\ln\pars{2} +
4\color{#c00000}{\int_{0}^{\pi/4}\ln\pars{\cos\pars{x}}\,\dd x}
\end{align}
A: One term is obvious from symmetry.  The other is not elementary: Maple gets
$$ \dfrac{\pi^2}{8} - \dfrac{\pi^3}{96} + \dfrac{\pi}{2} \ln(2) - 2 \;\text{Catalan}$$
A: We can write $$\frac{x^2 \arctan^2 (x) + \tan(x)}{1+x^2}= \frac{x^2 \arctan^2 (x)}{1+x^2}+\frac{\tan(x)}{1+x^2} $$
Let $f(x)=\frac{x^2 \arctan^2 (x)}{1+x^2} \quad g(x)=\frac{\tan(x)}{1+x^2}$
$f(x)$ is an even function because $f(x)=f(-x)$, $g(x)$ is odd. The domain of the integral is symmetric, so we have $$\int_{-1}^{1} g(x) dx=0$$
$$\int_{-1}^{1} f(x)dx   =  2\int_{0}^{1} f(x)dx$$ 
The whole integral is reduced to the solution of $$2\int_{0}^{1} \frac{x^2 \arctan^2 (x)}{1+x^2}dx$$
