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.

Here is a step-by-step approach: let $I$ denote the integral to be computed.


*

*Split $I$ into an integral on $(0,1)$ and an integral on $(1,+\infty)$.

*Deduce from this decomposition that $I=J+K$ where $$J=\int_0^1\frac{\log(1+x)}x\mathrm dx,\qquad K=\int_0^1\frac{-x\log(x)}{1+x^2}\mathrm dx.$$

*Integrate $K$ by parts, using the functions $u=\log(1+x^2)$ and $v=-\log x$, thus $$K=\frac12L,\qquad L=\int_0^1\frac{\log(1+x^2)}x\mathrm dx.$$

*Use the change of variable $x\to x^2$ in $L$ to deduce that $$L=\frac12J.$$

*Expand $\log(1+x)$ as an alternated series $$\log(1+x)=\sum_{n\geqslant1}(-1)^{n-1}\frac{x^n}n,$$ to deduce that $$J=\sum_{n\geqslant1}(-1)^{n-1}\left.\frac{x^n}{n^2}\right|_0^1=S,\qquad S=\sum_{n\geqslant1}\frac{(-1)^{n-1}}{n^2}.$$

*Note that $$S=\sum_{n\geqslant1}\frac{1}{n^2}-\sum_{n\geqslant1}\frac{2}{(2n)^2}=\frac12\zeta(2).$$

*Deduce from all this that $$I=\frac54J=\frac58\zeta(2).$$

*Conclude with $$\zeta(2)=\frac{\pi^2}6.$$

