Show that : $\displaystyle\int_0^{\infty}\frac{\ln(1+x^2)\operatorname{arc\,cot} x}{x}=\frac{π^3}{12}$ I need to prove this : 
$I=\displaystyle\int_0^{\infty}\frac{\ln(1+x^2)\operatorname{arc\mkern2mucot} x}{x}=\frac{π^3}{12}$
My try : 
$I\displaystyle\int_0^{1}\frac{\ln(1+x^2)\operatorname{arc\mkern2mucot} x}{x}$
$+\displaystyle\int_1^{\infty}\frac{\ln(1+x^2)\operatorname{arc\mkern2mucot} x}{x}$
$y=\frac{1}{x}$ so $dx=-\frac{dy}{y^2}$
$I=\displaystyle\int_0^{1}\frac{\ln(1+x^2)\operatorname{arc\mkern2mucot} x}{x}$
+$\displaystyle\int_0^{1}\frac{(\ln(1+x^2)-2\ln x)\arctan x}{x}$
Now I need use series of $\arctan x=\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{2k+1}$
But I don't know how I complete other integration !!
 A: The  key here is the identity:
$$\mathrm{arccot}x + \arctan x = \arctan \frac{1}{x} + \arctan x =  \frac{\pi}{2} \quad \quad \text{forall} \quad x>0$$
Thus,
\begin{align*}
\int_{0}^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x &= \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x + \int_{1}^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x} \, \mathrm{d}x \\ 
 &=\int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x + \int_{0}^{1} \frac{\ln \left ( 1+\frac{1}{x^2} \right ) \arctan x}{x} \, \mathrm{d}x \\ 
 &=\int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x + \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \arctan x}{x}\, \mathrm{d}x - \\
 &\quad \quad \quad - \int_0^1 \frac{2 \ln x \arctan x}{x} \, \mathrm{d}x  \\ 
 &= \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \left ( \mathrm{arccot} x + \arctan x \right )}{x} \, \mathrm{d}x  - \\
 & \quad \quad \quad - 2 \int_{0}^{1} \frac{\ln x \arctan x}{x} \, \mathrm{d}x \\ 
 &= \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \left ( \arctan \frac{1}{x} + \arctan x \right )}{x} \, \mathrm{d}x  - \\
 &\quad \quad \quad - 2 \int_{0}^{1} \frac{\ln x \arctan x}{x} \, \mathrm{d}x \\
 &=\frac{\pi}{2} \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{x} \, \mathrm{d}x + \frac{\pi^3}{16} \\
 &=  \frac{\pi^2}{24} \cdot \frac{\pi}{2} + \frac{\pi^3}{16} \\
 &= \frac{\pi^3}{12}
\end{align*}
The latter integrals are rather easy. If needed I can a add a calculation. 
Addendum:
For the second integral:
\begin{align*}
\int_{0}^{1} \frac{\ln x \arctan x}{x} \, \mathrm{d}x  &= \int_{0}^{1} \frac{\ln x}{x} \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n+1}}{2n+1} \, \mathrm{d}x \\ 
 &=\int_{0}^{1} \ln x \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{2n+1} \, \mathrm{d}x \\ 
 &=\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_{0}^{1}x^{2n} \ln x \, \mathrm{d}x \\ 
 &=-\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\ 
 &= -\beta(3) \\
 &=-\frac{\pi^3}{32}
\end{align*}
where $\beta$ is the Dirichlet Beta function. 
For the first integral:
\begin{align*}
\int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{x} \, \mathrm{d}x &= \int_{0}^{1} \frac{1}{x} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n}}{n} \, \mathrm{d}x \\ 
 &=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_{0}^{1} x^{2n-1} \, \mathrm{d}x \\ 
 &= \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} \, \mathrm{d}x  \\ 
 &= \frac{\eta(2)}{2}\\ 
 &= \frac{\left ( 1-2^{1-2} \right ) \zeta(2)}{2} \\
 &=\frac{\left ( 1-\frac{1}{2} \right ) \zeta(2)}{2} \\
 &=\frac{\zeta(2)}{4} \\
 &= \frac{\pi^2}{24} 
\end{align*}
where $\eta$ is the Dirichlet eta function and $\zeta$ is the Riemann zeta function. 
