How to show that $ \int^{\infty}_{0} \frac{\ln (1+x)}{x(x^2+1)} \ dx = \frac{5{\pi}^2}{48} $ without complex analysis? The Problem
I am trying to show that $ \displaystyle \int^{\infty}_{0} \frac{\ln (1+x)}{x(x^2+1)} \ dx = \frac{5{\pi}^2}{48}$ 
My attempt
I've tried substituting $x=\tan\theta$, and then using the substitution $u=1 + \tan \theta $ which gives:
$ \displaystyle \int^{\infty}_{1} \frac{\ln u}{(u-1)(u^2-2u+2)} \ du $ , however I am unable to evaluate this.

We have:
$$I=\int_{0}^{+\infty}\frac{\log(u+1)}{u^3+u}\,du =\int_{0}^{+\infty}\int_{0}^{1}\frac{1}{(u^2+1)(1+uv)}\,dv\,du$$
and by exchanging the order of integration, then setting $v=\sqrt{w}$:
$$ I = \int_{0}^{1}\frac{\pi +2v\log v}{2+2v^2}\,dv =\frac{\pi^2}{8}+\int_{0}^{1}\frac{v\log v}{1+v^2}\,dv=\frac{\pi^2}{8}+\frac{1}{4}\int_{0}^{1}\frac{\log w}{1+w}\,dw$$
so:
$$ I = \frac{\pi^2}{8}-\frac{\pi^2}{48} = \color{red}{\frac{5\pi^2}{48}}. $$
