Integral $\int_0^{\infty}\frac{\operatorname{arctg}^2x}{x^2}dx$ What's the way to find the following integral $\int_0^{\infty}\frac{\operatorname{arctg}^2x}{x^2}\,dx$? I know the answer is $\pi \ln2$.
I tried to integrate by parts and got
$$2\int_0^{\infty}\frac{\operatorname{arctg}x}{x(1+x^2)}\,dx.$$
I don't know what to do next, maybe another integration by parts could help or this was a wrong path at all? 
 A: Since $\displaystyle \frac{\arctan{x}}{x} = \int_0^1 \frac{1}{1+x^2y^2}\, \mathrm{dy}$, we have: 
$$\begin{aligned} I & = \int_0^{\infty} \frac{\arctan{x}}{x(1+x^2)}\,\mathrm{dx}  \\& = \int_0^{\infty} \int_0^1 \frac{1}{(1+x^2)(1+x^2y^2)}\,\mathrm{dy}\,\mathrm{dx} \\& = \int_0^{1} \int_0^\infty \frac{1}{(1+x^2)(1+x^2y^2)}\,\mathrm{dx}\,\mathrm{dy} \\& =\frac{\pi}{2} \int_0^{1}  \frac{1}{1+y}\,\mathrm{dy} \\& =\pi \log \sqrt{2}. \end{aligned}   $$
A: Let
$$I(a,b)=\int_0^{\infty}\frac{\arctan(ax)\arctan(bx)}{x^2}\,dx.$$
Clearly $I(a,0)=I(0,b)=I(0,0)=0$ and $I(1,1)=0$. Note
\begin{eqnarray}
\frac{\partial^2I(a,b)}{\partial a\partial b}&=&\int_0^{\infty}\frac1{(1+a^2x^2)(1+b^2x^2)}\,dx\\
&=&\int_0^{\infty}\bigg(\frac{a^2}{a^2-b^2}\frac{1}{1+a^2x^2}-\frac{b^2}{a^2-b^2}\frac{1}{1+b^2x^2}\bigg)\,dx\\
&=&\frac{a^2}{a^2-b^2}\frac{\pi}{2a}-\frac{b^2}{a^2-b^2}\frac{\pi}{2b}\\
&=&\frac{\pi}{2(a+b)}.
\end{eqnarray}
So
$$ I=\int_0^1\int_0^1\frac{\pi}{2(a+b)}dadb=\pi\ln2. $$
A: As you said :
\begin{aligned}I=\int_{0}^{+\infty}{\frac{\arctan^{2}{x}}{x^{2}}\,\mathrm{d}x}&=2\int_{0}^{+\infty}{\frac{\arctan{x}}{x\left(1+x^{2}\right)}\,\mathrm{d}x}\end{aligned}
Using the substitution $\small \left\lbrace\begin{aligned}y&=\frac{1}{x}\\ \mathrm{d}x&=-\frac{\mathrm{d}x}{x^{2}}\end{aligned}\right. $, we get : \begin{aligned} I=\int_{0}^{+\infty}{\frac{2y}{1+y^{2}}\arctan{\left(\frac{1}{y}\right)}\,\mathrm{d}y}&=\left[\ln{\left(1+y^{2}\right)}\arctan{\left(\frac{1}{y}\right)}\right]_{0}^{+\infty}+\int_{0}^{+\infty}{\frac{\ln{\left(1+y^{2}\right)}}{1+y^{2}}\,\mathrm{d}y}\\ &=\int_{0}^{+\infty}{\frac{\ln{\left(1+y^{2}\right)}}{1+y^{2}}\,\mathrm{d}y} \end{aligned}
Using another substitution $\small \left\lbrace\begin{aligned}y&=\tan{x}\\ \mathrm{d}x&=\frac{\mathrm{d}y}{1+y^{2}}\end{aligned}\right. $, we get : \begin{aligned}I=-2\int_{0}^{\frac{\pi}{2}}{\ln{\left(\cos{x}\right)}\,\mathrm{d}x}=-2\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{x}\right)}\,\mathrm{d}x}\end{aligned}
And since :
\begin{aligned} \int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{x}\right)}\,\mathrm{d}x}=2\int_{0}^{\frac{\pi}{4}}{\ln{\left(\sin{\left(2t\right)}\right)}\,\mathrm{d}t}&=2\int_{0}^{\frac{\pi}{4}}{\ln{\left(2\sin{t}\cos{t}\right)}\,\mathrm{d}t}\\&=\frac{\pi}{2}\ln{2}+2\int_{0}^{\frac{\pi}{4}}{\ln{\left(\sin{t}\right)}\,\mathrm{d}t}+2\int_{0}^{\frac{\pi}{4}}{\ln{\left(\cos{t}\right)}\,\mathrm{d}t}\\ &=\frac{\pi}{2}\ln{2}+2\int_{0}^{\frac{\pi}{4}}{\ln{\left(\sin{t}\right)}\,\mathrm{d}t}+2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\ln{\left(\sin{u}\right)}\,\mathrm{d}u} \\ \int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{t}\right)}\,\mathrm{d}t}&=\frac{\pi}{2}\ln{2}+2\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{t}\right)}\,\mathrm{d}t}\\ \iff \int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{t}\right)}\,\mathrm{d}t}&=-\frac{\pi}{2}\ln{2}\end{aligned}
We get that : $$ I=\pi\ln{2} $$
