TITLE: A tricky integral
QUESTION [17 upvotes]: 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 ...

REPLY [21 votes]: 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.