A tricky integral I'm trying to find the exact value of $$\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{\arctan{(x^2)} }{1+x^2} \, dx$$
Ostensibly, I'd want to use this: $$\frac{d}{dx}\arctan{(x)}=\frac{1}{1+x^2}$$ 
But either I'm missing something, or this doesn't work out nicely ...
 A: Use the substitution $x=\tan\theta$ to obtain:
$$I=\int_{\pi/6}^{\pi/3} \arctan(\tan^2\theta)\,d\theta $$
Since
$$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$$
We get:
$$I=\int_{\pi/6}^{\pi/3} \arctan(\cot^2\theta)\,d\theta=\int_{\pi/6}^{\pi/3} \text{arccot}(\tan^2\theta)\,d\theta$$
Add the two to get:
$$2I=\int_{\pi/6}^{\pi/3} \frac{\pi}{2}\,d\theta \Rightarrow I=\frac{\pi^2}{24}$$
....which well agrees with Wolfram Alpha.
I hope this helps.
A: \begin{align}
u & = \frac 1 x \\[8pt]
du & = \frac{-dx}{x^2} \\[8pt]
\frac{-du}{u^2} & = dx
\end{align}
\begin{align}
I = & \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{\arctan(x^2)}{1+x^2} \, dx \\[10pt]
= & \int_\sqrt{3}^{1/\sqrt{3}} \frac{\arctan\frac{1}{u^2}}{1+\left(\frac{1}{u^2}\right)} \left(\frac{-du}{u^2}\right) \\[10pt]
= & \int_{1/\sqrt{3}}^\sqrt{3} \frac{\arctan\left(\frac{1}{u^2}\right)}{u^2+1} \,du \\[10pt]
= & \int_{1/\sqrt{3}}^\sqrt{3} \frac{\frac\pi2 - \arctan(u^2)}{u^2+1} \,du \\[10pt]
= & \frac \pi 2 \int_{1/\sqrt{3}}^\sqrt{3} \frac{du}{1+u^2} - \int_{1/\sqrt{3}}^\sqrt{3} \frac{\arctan(u^2)}{1+u^2} \,du \\[10pt]
= & \frac \pi 2 \int_{1/\sqrt{3}}^\sqrt{3} \frac{du}{1+u^2} - I.
\end{align}
So we have
$$
I = \left(\int\cdots\cdots\cdots\right) - I,
$$
whence
$$
2I = \left(\int\cdots\cdots\cdots\right).
$$
That last integral is routine.  Remember that $\arctan\sqrt{3} = \pi/3$ and $\arctan(1/\sqrt{3})= \pi/6$.
