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)?
$$F(n)=\int_0^1 \frac{\log^n(1+x)}{x} \ dx, \space n\in \mathbb{N}$$  
 A: Different approach:
We have 
$$\ln^2(1+x)=2\sum_{n=1}^\infty\frac{H_n}{n+1}(-x)^{n+1}$$
Divide by $x$ then integrate to get 
\begin{align}
\int_0^1\frac{\ln^2(1+x)}{x}\ dx&=2\sum_{n=1}^\infty\frac{(-1)^{n+1}H_n}{n+1}\int_0^1x^n\ dx\\
&=2\sum_{n=1}^\infty\frac{(-1)^{n+1}H_n}{(n+1)^2}\\
&=2\sum_{n=1}^\infty\frac{(-1)^{n}H_{n-1}}{n^2}\\
&=2\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^2}-2\sum_{n=1}^\infty\frac{(-1)^{n}}{n^3}\\
&=2\left(-\frac58\zeta(3)\right)-2\operatorname{Li}_3(-1)\\
&=-\frac54\zeta(3)-2\left(-\frac34\zeta(3)\right)\\
&=\boxed{\frac14\zeta(3)}
\end{align}

Note:
We have the generating identity

$$\sum_{n=1}^\infty x^n\frac{H_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$

and by setting $x=-1$ and considering only the real parts we have 
$$\Re\sum_{n=1}^\infty (-1)^n\frac{H_n}{n^2}=\operatorname{Li}_3(-1)-\Re\operatorname{Li}_3(2)+\Re\ln2\operatorname{Li}_2(2)+\frac12\underbrace{\Re\ln(-1)\ln^22}_{0}+\zeta(3)\tag{1}$$
Using the trilogarithmic identity 
$$\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)+\operatorname{Li}_3\left(\frac{x-1}{x}\right)=\frac16\ln^3x+\zeta(2)\ln x-\frac12\ln^2x\ln(1-x)+\zeta(3)$$
set $x=-1$ and take the real parts to have
$$ \boxed{\Re\operatorname{Li}_3(2)=\frac78\zeta(3)+\frac32\ln2\zeta(2)}$$
also Landen's identity gives
$$ \boxed{\Re\operatorname{Li}_2(2)=\frac32\zeta(2)}$$
Plugging the boxed results along with $\operatorname{Li}_3(-1)=-\frac34\zeta(3)$ in (1) we have 

$$\Re\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^2}=-\frac58\zeta(3)$$

A: You can find a nice generalization for $\int_0^1\frac{\ln^n(1+x)}{x}dx$ in lemma $2.2$ in this article and I am going to type it here with little more details.
Start with subbing $\frac{1}{1+x}=y$
$$I_n=\int_0^1\frac{\ln^n(1+x)}{x}dx=(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{y(1-y)}dy$$
$$=(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{y}dy+(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{1-y}dy$$
$$=(-1)^n\left[(-1)^n\frac{\ln^{n+1}(2)}{n+1}\right]+(-1)^n\int_{0}^1\frac{\ln^n(y)}{1-y}dy-(-1)^n\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy$$
$$=(-1)^n\left[(-1)^n\frac{\ln^{n+1}(2)}{n+1}\right]+(-1)^n\left[(-1)^n n!\zeta(n+1)\right]-(-1)^n\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy$$
$$=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)-(-1)^n\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy\tag1$$
By using 
$$(x+y)^n=\sum_{k=0}^n{n\choose k}x^{n-k}y^k$$
or
$$(x-y)^n=(-1)^n(y-x)^n=(-1)^n \sum_{k=0}^n{n\choose k}y^{n-k}(-x)^k=\sum_{k=0}^n{n\choose k}(-y)^{n-k}x^k\tag2$$
we get
$$\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy\overset{2y=x}{=}-\int_0^1\frac{(\ln(x)-\ln(2))^n}{2-x}dx$$
$$\overset{(2)}{=}-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}\left(\int_0^1\frac{\ln^k(x)}{2-x}dx\right)$$
$$=-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}\left(\sum_{i=1}^\infty\frac1{2^i} \int_0^1 x^{i-1}\ln^k(x)dx\right)$$
$$=-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}\left((-1)^k k!\sum_{i=1}^\infty\frac1{2^i i^{k+1}}\right)$$
$$=-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}(-1)^k k!\operatorname{Li}_{k+1}\left(\frac12\right)\tag3$$
Plug $(3)$ in $(1)$ we get 

$$I_n=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$
  or
  $$(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{y(1-y)}dy=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$

A: 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)
A: Squaring the series for $\log(1+x)$ yields
$$
\log(1+x)^2=\sum_{k=2}^\infty\sum_{j=1}^{k-1}\frac{(-1)^kx^k}{j(k-j)}
$$
Dividing by $x$ and integrating gives
$$
\begin{align}
\int_0^1\frac{\log(1+x)^2}{x}\mathrm{d}x
&=\sum_{k=2}^\infty\sum_{j=1}^{k-1}\frac{(-1)^k}{jk(k-j)}\\
&=\sum_{j=1}^\infty\sum_{k=j+1}^\infty\frac{(-1)^k}{jk(k-j)}\\
&=\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{j+k}}{jk(j+k)}\\[9pt]
&=\frac{\zeta(3)}{4}
\end{align}
$$
Using $(5)$ from this answer:
$$
\sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n
=-\frac34\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{n+k}}{(n+k)kn}
$$
and $(6)$ from the same answer:
$$
-\frac58\zeta(3)
=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n
$$
we get
$$
\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{j+k}}{jk(j+k)}
=\frac{\zeta(3)}{4}
$$
A: Using the algebraic identity 
$$b^2=\frac12(a-b)^2+\frac12(a+b)^2-a^2$$
let $a=\ln(1-x)$ and $b=\ln(1+x)$ we have 
$$\int_0^1\frac{\ln^2(1+x)}{x}\ dx=\frac12\underbrace{\int_0^1\frac{\ln^2\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}=y}+\frac12\underbrace{\int_0^1\frac{\ln^2(1-x^2)}{x}\ dx}_{1-x^2=y}-\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ dx}_{1-x=y}\\=\int_0^1\frac{\ln^2y}{1-y^2}\ dy+\frac14\int_0^1\frac{\ln^2y}{1-y}\ dy-\int_0^1\frac{\ln^2y}{1-y}\ dy\\=\frac12\int_0^1\frac{\ln^2y}{1+y}\ dy-\frac14\int_0^1\frac{\ln^2y}{1-y}\ dy=\frac12\left(\frac32\zeta(3)\right)-\frac14(2\zeta(3))=\boxed{\frac14\zeta(3)}$$
A: Here is a solution by finding the closed form of $\int \frac{\ln^2(1-x)}{x}dx$ then letting $x\mapsto -x$:
$$\int \frac{\ln^2(1-x)}{x}dx=\int \frac{\ln(1-x)\ln(1-x)}{x}dx\overset{IBP}{=}-\operatorname{Li}_2(x)\ln(1-x)-\int\frac{\operatorname{Li}_2(x)}{1-x}dx$$
For the last integral, set $1-x=y$ then use the reflection formula:
$$\operatorname{Li}_2(1-y)=\zeta(2)-\ln(y)\ln(1-y)-\operatorname{Li}_2(y)$$
We obtain that
$$\int\frac{\operatorname{Li}_2(x)}{1-x}dx=-\int\frac{\operatorname{Li}_2(1-y)}{y}dy$$
$$=-\zeta(2)\int\frac{dy}y+\int\frac{\ln(y)\ln(1-y)}{y}dy+\int\frac{\operatorname{Li}_2(y)}{y}dy$$
$$=-\zeta(2)\ln(y)+\left[-\operatorname{Li}_2(y)\ln(y)+\int\frac{\operatorname{Li}_2(y)}{y}dy\right]+\int\frac{\operatorname{Li}_2(y)}{y}dy$$
$$=-\zeta(2)\ln(y)-\operatorname{Li}_2(y)\ln(y)+2\operatorname{Li}_3(y)$$
$$=-\zeta(2)\ln(1-x)-\operatorname{Li}_2(1-x)\ln(1-x)+2\operatorname{Li}_3(1-x)$$
Then
$$\int\frac{\ln^2(1-x)}{x}dx=\ln(1-x)\left[\operatorname{Li}_2(1-x)-\operatorname{Li}_2(x)+\zeta(2)\right]-2\operatorname{Li}_3(1-x)$$
Now consider the integral boundaries $(0,a)$, 
$$\int_0^a\frac{\ln^2(1-x)}{x}dx=\ln(1-a)\left[\operatorname{Li}_2(1-a)-\operatorname{Li}_2(a)+\zeta(2)\right]-2\operatorname{Li}_3(1-a)+2\zeta(3)$$
Therefore
$$\int_0^1\frac{\ln^2(1+x)}{x}dx\overset{x\mapsto -x}{=}\int_0^{-1}\frac{\ln^2(1-x)}{x}dx$$
$$=\ln(2)\left[\operatorname{Li}_2(2)-\operatorname{Li}_2(-1)+\zeta(2)\right]-2\operatorname{Li}_3(2)+2\zeta(3)$$
substitute $\Re\operatorname{Li}_2(2)=\frac32\zeta(2)$ and $\Re\operatorname{Li}_3(2)=\frac78\zeta(3)+\frac32\ln2\zeta(2)$, the closed form follows.
