Ways to prove $ \int_0^1 \frac{\ln^2(1+x)}{x}dx = \frac{\zeta(3)}{4}$? I am wondering if we can show in a simple way that
$$
I=\int_0^1 \frac{\ln^2(1+x)}{x}dx = \int_1^2 \frac{\ln^2(t)}{t-1}dt = \frac{\zeta(3)}{4}.
$$
Because the end result is very simple, I suspect that there might be a fast way to prove this.
Can you prove it without using polylog identities? Complex analysis is allowed.
It may be easier to show the equivalent identity
$$
\sum_{k=1}^\infty \frac{(-1)^k H_k}{k^2} = -\frac{5 \zeta (3)}{8}
$$
I know you can do that one with the generating function of the harmonic numbers, but that gives a nasty expression with polylogs which I would like to avoid.
 A: Let us denote $I_{\pm}=\displaystyle \int_{0}^1\frac{\ln^2(1\pm x)}{x}dx$. We will express $I_+$ in terms of $I_-$, which is itself related to the standard integral representation of $\zeta(z)$ by the change of variables $x=1-e^{-t}$:
$$I_-=\int_0^{\infty}\frac{t^2dt}{e^{t}-1}=2\zeta(3).$$
Indeed, we have
 \begin{align} \int_0^1\frac{\ln^2\frac{1+x}{1-x}}{x}dx=\int_0^{\infty}\frac{16t^2}{2\sinh 2t}dt&=\int_0^{\infty}16t^2\left(\frac{1}{e^{2t}-1}-\frac{1}{e^{4t}-1}\right)dt=\frac74 I_- \tag{1}
 \end{align}
where the first equality is obtained by setting $x=\tanh t$. Also, it is easy to show ($x^2\to x$) that
$$\int_{0}^1\frac{\ln^2(1-x^2)}{x}dx=\frac12I_-. \tag{2}$$
Summing  (1) and (2), one finds that
$ 2I_+ +2I_-=\left(\frac74+\frac12\right)I_-$, and hence $\displaystyle I_+=\frac{I_-}{8}=\frac{\zeta(3)}4$.
A: Using this answer which shows that
$$
\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}H_n=\frac58\zeta(3)
$$
and the series
$$
\frac{\log(1+x)}{1+x}=\sum_{k=1}^\infty(-1)^{k-1}H_kx^k
$$
we get
$$
\begin{align}
\int_0^1\frac{\log(1+x)^2}{x}\mathrm{d}x
&=\int_0^1\log(1+x)^2\,\mathrm{d}\log(x)\\
&=-2\int_0^1\frac{\log(1+x)\log(x)}{1+x}\,\mathrm{d}x\\
&=-2\int_0^1\sum_{k=1}^\infty(-1)^{k-1}H_kx^k\log(x)\,\mathrm{d}x\\
&=2\sum_{k=1}^\infty\frac{(-1)^{k-1}H_k}{(k+1)^2}\\
&=2\sum_{k=1}^\infty\frac{(-1)^{k-1}H_{k+1}}{(k+1)^2}-2\sum_{k=1}^\infty\frac{(-1)^{k-1}}{(k+1)^3}\\
&=2\left(\frac34\zeta(3)-\sum_{k=1}^\infty\frac{(-1)^{k-1}H_k}{k^2}\right)\\
&=\frac{\zeta(3)}4
\end{align}
$$
A: Hmm, i don't know if this answer fulfills the requirement to be a fast way, but it is relatively straightforward:
1.) Use the sub $1+x=e^y$
The integral is now 
$$
\int_0^{\log(2)}\frac{y^2}{1-e^{-y}}dy
$$
2.) By help of geometric series we obtain
$$
\sum_{n=0}^{\infty}\int_{0}^{\log(2)}y^2e^{-ny}dy
$$ 
3.) Seperating the $n=0$  term and doing the trivial integrations we obtain
$$
-\left(\log^2(2)\sum_{n=1}^{\infty}\frac{1}{n2^n}+2\log(2)\sum_{n=1}^{\infty}\frac{1}{n^22^n}+2\sum_{n=1}^{\infty}\frac{1}{n^32^n}-2\sum_{n=1}^{\infty}\frac{1}{n^3}\right)+\frac{1}{3}\log^3(2)
$$
4.) Remembering the definition of Polylog $\text{Li}_s(z)=\sum_{n=1}^{\infty}\frac{z^n}{n^s}$ we can now look up the values $\text{Li}_{1}(1/2),\text{Li}_{2}(1/2),\text{Li}_{3}(1/2)$ in some table and put everything together to obtain (magic!)
$$
\frac{\zeta(3)}{4}
$$
A: Here is a particularly efficient way to get to your Euler sum. 
In this post I show that
$$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n}.$$
Replacing $x $ with $-x$ gives
$$\ln^2 (1 + x) = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1} x^n}{n}.$$
So if we replace the term $\ln^2 (1 + x)$ with its above Maclaurin series expansion the integral becomes
$$\int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1}}{n} \int_0^1 x^{n - 1} \, dx = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1}}{n^2}.$$
Making use of the following property for harmonic numbers
$$H_n = H_{n - 1} + \frac{1}{n},$$
the integral can be expressed as
$$\int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_n}{n^2} - 2 \sum_{n = 2}^\infty \frac{(-1)^n}{n^3} = 2 \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} - 2 \sum_{n = 1}^\infty \frac{(-1)^n}{n^3}.$$
For the sums, as you note
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} = -\frac{5}{8} \zeta (3),$$
and
$$\sum_{n = 1}^\infty \frac{(-1)^n}{n^3} = - \sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^3} = - \eta (3) = -(1 - 2^{1-3}) \zeta (3) = -\frac{3}{4} \zeta (3),$$
where $\eta (s)$ is the Dirichlet eta function, one finally has
$$\int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx = -\frac{5}{4} \zeta (3) + \frac{3}{2} \zeta (3) = \frac{1}{4} \zeta (3),$$
as expected. 
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large\int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x}\
\stackrel{\dsc{1 + x}\ \mapsto\ \dsc{x}}{=}\
\int_{1}^{2}{\ln^{2}\pars{x} \over x - 1}\,\dd x\
\stackrel{\dsc{x}\ \mapsto\ \dsc{1 \over x}}{=}\
\int_{1}^{1/2}{\ln^{2}\pars{1/x} \over 1/x - 1}\,\pars{-\,{\dd x \over x^{2}}}
\\[5mm]&=\int_{1/2}^{1}\ {\ln^{2}\pars{x} \over x\pars{1 - x}}\,\dd x
=\int_{1/2}^{1}\ {\ln^{2}\pars{x} \over x}\,\dd x
+\int_{1/2}^{1}\ {\ln^{2}\pars{x} \over 1 - x}\,\dd x
\\[5mm]&={1 \over 3}\,\ln^{3}\pars{2}
-\left.\vphantom{\LARGE A}\ln\pars{1 - x}\ln^{2}\pars{x}\right\vert_{1/2}^{1}
+\int_{1/2}^{1}\ln\pars{1 - x}\bracks{2\ln\pars{x}\,{1 \over x}}\,\dd x
\\[5mm]&=-\,{2 \over 3}\,\ln^{3}\pars{2} - 2\int_{1/2}^{1}\Li{2}'\pars{x}\ln\pars{x}\,\dd x
\end{align}
where $\Li{\rm s}$ is a
PolyLogarithm Function. We already used the identity
$\ds{\Li{\rm s}'\pars{t}=
 {\Li{{\rm s} - 1}\pars{t} \over t}}$ with $\Li{1}\pars{t}=-\ln\pars{1 - t}$.

Then,
\begin{align}
&\color{#66f}{\large\int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x}
=-\,{2 \over 3}\,\ln^{3}\pars{2} - 2\Li{2}\pars{\half}\ln\pars{2}
+2\int_{1/2}^{1}\Li{2}\pars{x}\,{1 \over x}\,\dd x
\\[5mm]&=-\,{2 \over 3}\,\ln^{3}\pars{2} - 2\Li{2}\pars{\half}\ln\pars{2}
+2\int_{1/2}^{1}\Li{3}'\pars{x}\,\dd x
\\[5mm]&=-\,{2 \over 3}\,\ln^{3}\pars{2} - 2\Li{2}\pars{\half}\ln\pars{2}
+2\ \overbrace{\Li{3}\pars{1}}^{\dsc{\zeta\pars{3}}}\ - 2\Li{3}\pars{\half}
\\[1cm]&=-\,{2 \over 3}\,\ln^{3}\pars{2}
-2\ \overbrace{\bracks{{1 \over 12}\,\pi^{2} - \half\,\ln^{2}\pars{2}}}
^{\ds{=\ \dsc{\Li{2}\pars{\half}}}}\ \ln\pars{2} + 2\zeta\pars{3}
\\[5mm]&\phantom{=}- 2\ \overbrace{%
\bracks{{1 \over 6}\,\ln^{3}\pars{2} - {1 \over 12}\,\pi^{2}\ln\pars{2} + {7 \over 8}\,\zeta\pars{3}}}^{\ds{=\ \dsc{\Li{3}\pars{\half}}}}
\ =\ \color{#66f}{\large{1 \over 4}\,\zeta\pars{3}}
\end{align}


$\ds{\Li{2}\pars{\half}}$ and $\ds{\Li{3}\pars{\half}}$ are well known values ( a few ones !!! ) and they are given elsewhere.

A: Using the fact that
$$\ln^2(1+x)=\frac12\ln^2\left(\frac{1-x}{1+x}\right)+\frac12\ln^2(1-x^2)-\ln^2(1-x)$$
we have
$$\int_0^1\frac{\ln^2(1+x)}{x}dx=\frac12\underbrace{\int_0^1\frac{\ln^2\left(\frac{1-x}{1+x}\right)}{x}dx}_{(1-x)/(1+x)\to x}+\frac12\underbrace{\int_0^1\frac{\ln^2(1-x^2)}{x}dx}_{1-x^2\to x}-\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}dx}_{1-x\to x}$$
$$=\int_0^1\frac{\ln^2(x)}{1-x^2}dx+\frac14\int_0^1\frac{\ln^2(x)}{1-x}dx-\int_0^1\frac{\ln^2(x)}{1-x}dx.$$
Since $$\frac1{1-x^2}-\frac1{1-x}=\frac{x}{1-x^2},$$ we have
$$\int_0^1\frac{\ln^2(1+x)}{x}dx=\frac14\int_0^1\frac{\ln^2(x)}{1-x}dx-\underbrace{\int_0^1\frac{x\ln^2(x)}{1-x^2}dx}_{x^2\to x}$$
$$=\frac18\int_0^1\frac{\ln^2(x)}{1-x}dx=\frac14\zeta(3).$$
Note:
$$\int_0^1\frac{\ln^a(x)}{1-x}dx=\sum_{n=1}^\infty\int_0^1 x^{n-1}\ln^a(x)dx=\sum_{n=1}^\infty\frac{(-1)^aa!}{n^{a+1}}=(-1)^aa!\zeta(a+1).$$
