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 ...

Solution 1:

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.

Solution 2:

\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 $\arctan(1/\sqrt{3})= \pi/6$.