How to prove that $\int_{0}^{\infty}{\ln[x(1+x)]\ln\left(x\over 1+x\right)\over (x+2)^2}\mathrm dx={1\over 2}\ln^2(2)?$ Log integral
$$\int_{0}^{\infty}{\ln[x(1+x)]\ln\left(x\over 1+x\right)\over (x+2)^2}\mathrm dx={1\over 2}\ln^2(2)\tag1$$
Making an attempt:
Following from my previous post making $u={x\over 1+x}$ it doesn't worked, so I try applying binomial series
$(1)$ becomes 
$$\sum_{n=0}^{\infty}(-1)^n\cdot{n+1\over 2^n}\int_{0}^{\infty}[x^n\ln^2(x)-x^n\ln^2(1+x)]\mathrm dx=2\ln^2(2)\tag2$$
Applying IBP to these indefinite integrals
$$\int x^n\ln^2(x)\mathrm dx={x^{n+1}\over n+1}\cdot \ln^2(x)-x^2\cdot{2\ln x-1\over 2(n+1)}+C\tag3$$
$$\int x^n\ln^2(1+x)\mathrm dx={x^{n+1}\over n+1}\cdot \ln^2(1+x)-{2\over n+1}\int\color{red}{{x^{n+1}\over 1+x}\ln(1+x)\mathrm dx}\tag4$$
$$\color{red}{\int{x^{n+1}\over 1+x}\ln(1+x)}\mathrm dx={x^{n+1}\over 2}\ln^2(1+x)-{n+1\over 2}\int x^n\ln^2(1+x)\mathrm dx\tag5$$
If I put $(5)$ into $(4)$ we get zero! What I am doing here doesn't seem to be working. 
How can we go about to tackle $(1)?$
 A: $$\int_{0}^{+\infty}\frac{\log^2(x)-\log^2(x+1)}{(x+2)^2}\,dx = I_1-I_2 = \int_{0}^{+\infty}\frac{\log^2(x)}{(x+2)^2}\,dx-\int_{1}^{+\infty}\frac{\log^2(x)}{(x+1)^2}\,dx$$
is quite simple to compute. By integration by parts $I_2 = 2\zeta(2)=\frac{\pi^2}{3}$, while through the same technique we get that $I_1$ depends on $\text{Li}_2\left(\frac{1}{2}\right)$. The dilogarithm reflection formula hence settles the question.
A: On the path of Jack D'Aurizio,
$I_1=\displaystyle \int_0^{+\infty} \dfrac{(\ln(x))^2}{(x+2)^2}dx$
In $I_1$ perform the change of variable $y=\dfrac{x}{2}$,
$\begin{align} I_1&=2\int_0^{+\infty} \dfrac{(\ln(2x))^2}{(2x+2)^2}dx\\
&=\dfrac{1}{2}\int_0^{+\infty} \dfrac{(\ln 2+\ln x)^2}{(1+x)^2}dx\\
&=\dfrac{1}{2}(\ln 2)^2\int_0^{+\infty}\dfrac{1}{(1+x)^2}dx+\ln 2\int_0^{+\infty}\dfrac{\ln x}{(1+x)^2}dx+\dfrac{1}{2}\int_0^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx\\
&=\dfrac{1}{2}(\ln 2)^2\left[-\dfrac{1}{1+x}\right]_0^{+\infty}+\dfrac{1}{2}\int_0^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx+\ln 2\int_0^1 \dfrac{\ln x}{(1+x)^2}dx+\\
&\ln 2\int_1^{+\infty} \dfrac{\ln x}{(1+x)^2}dx\\
\end{align}$
In the latter integral perform the change of variable $y=\dfrac{1}{x}$,
$\begin{align}I_1&=\dfrac{1}{2}(\ln 2)^2+\dfrac{1}{2}\int_0^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx+\ln 2\int_0^1 \dfrac{\ln x}{(1+x)^2}dx-\ln 2\int_0^1 \dfrac{\ln x}{(1+x)^2}dx\\
&=\dfrac{1}{2}(\ln 2)^2+\dfrac{1}{2}\int_0^{1}\dfrac{(\ln x)^2}{(1+x)^2}dx+\dfrac{1}{2}\int_1^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx\\
\end{align}$
In the latter line, in the first integral perform the change of variable $y=\dfrac{1}{x}$,
$\begin{align}I_1&=\dfrac{1}{2}(\ln 2)^2+\dfrac{1}{2}\int_1^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx+\dfrac{1}{2}\int_1^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx\\
&=\dfrac{1}{2}(\ln 2)^2+I_2
\end{align}$
Therefore,
$\boxed{I_1-I_2=\dfrac{1}{2}(\ln 2)^2}$
