Evaluation of $\int_0^1 \frac{\log^2(1+x)}{x} \ dx$ One of the ways to approach it lies in the area of the dilogarithm, but is it possible to evaluate it
by other means of the real analysis (without using dilogarithm)?
$$\int_0^1 \frac{\log^2(1+x)}{x} \ dx$$  
EDIT: maybe you're aware of some easy way to do that. I'd appreciate it!
Some words on the generalization case (by means of the real analysis again)?
$$F(n)=\int_0^1 \frac{\log^n(1+x)}{x} \ dx, \space n\in \mathbb{N}$$  

On the path of Felix Marin,
\begin{align}J&=\int_0^1 \frac{\ln(1+x)^2}{x}\\
&\overset{y=\frac{1}{1+x}}=\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{x(1-x)}\,dx\\
&=\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{x}\,dx+\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{1-x}\,dx\\
&=\frac{1}{3}\left(\ln^3 (1)-\ln^3\left(\frac{1}{2}\right)\right)+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^{\frac{1}{2}} \frac{\ln^2 x}{1-x}\,dx\\
&=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^{\frac{1}{2}} \frac{\ln^2 x}{1-x}\,dx\\
&\overset{y=\frac{x}{1-x},\text{the 2nd integral}}=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2\left(\frac{x}{1+x}\right)}{1+x}\,dx\\
&=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2 x}{1+x}\,dx-\int_0^1\frac{\ln^2 (1+x)}{1+x}\,dx+2\int_0^1\frac{\ln(1+x)\ln x}{1+x}\,dx\\
&\overset{IBP}=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2 x}{1+x}\,dx-\int_0^1\frac{\ln^2 (1+x)}{1+x}\,dx-J\\
&=\frac{1}{3}\ln^3 2+\int_0^1 \frac{2x\ln^2 x}{1-x}\,dx-\frac{1}{3}\ln^3 2-J\\
&\overset{y=x^2}=\frac{1}{4}\int_0^1 \frac{\ln^2 x}{1-x}\,dx-J\\
J&=\frac{1}{8}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\
&=\frac{1}{8}\times 2\zeta(3)\\
&=\boxed{\frac{1}{4}\zeta(3)}
\end{align}
NB: i assume that, \begin{align}\int_0^1 \frac{\ln^2 x}{1-x}\,dx=2\zeta(3)\end{align}
(proof: Taylor expansion)
