About the integral $\int_{0}^{1}\frac{\log(x)\log^2(1+x)}{x}\,dx$ I came across the following Integral and have been completely stumped by it.

$$\large\int_{0}^{1}\dfrac{\log(x)\log^2(1+x)}{x}dx$$

I'm extremely sorry, but the only thing I noticed was that the limits of the Integral were similar to the Beta function. I also got a hint that solving it would require the Polylogarithm, Gamma and the Riemann Zeta Functions.
$$$$Would it be possible to solve this without using Complex Methods (I haven't learnt them yet) unless absolutely necessary? Any help on this Integral would be greatly appreciated. Many, many thanks in advance!
EDIT: From the comments given below byDavid H Sir, and Alex S Sir, the Integral becomes: $$\int_0^1 \dfrac{\ln^2(1+x)\ln(x)}{x}dx$$
Just an observation: This is strikingly similar to the Beta Function. Also, if we consider $$\int^1_0 (1+x)^ax^bdx$$ and differentiate twice with respect to $a$ and once with respect to $b$ and set $a=b=0$, we get the above Integral (except the $x$ in the denominator).
I'm not sure, but I think taking series representations of the Integrals would be of help, especially since the closed form includes the Riemann Zeta and the Polylogarithm functions.
 A: Here is a solution without using Euler sums
By integration by parts we have 
$$\int_0^1\frac{\ln x\ln^2(1+x)}{x}\ dx=-\int_0^1\frac{\ln^2x\ln(1+x)}{1+x}\ dx=-I$$
For $I$, start with the algebraic identity 
$$a^2b=\frac13a^3-\frac13b^3+ab^2-\frac13(a-b)^3$$
where if we set $a=\ln x$ and $b=\ln(1+x)$ we have 
$$I=\int_0^1\frac{\ln^2x\ln(1+x)}{1+x}\ dx$$
$$=\frac13\underbrace{\int_0^1\frac{\ln^3x}{1+x}\ dx}_{I_1}-\frac13\underbrace{\int_0^1\frac{\ln^3(1+x)}{1+x}\ dx}_{I_2}+\underbrace{\int_0^1\frac{\ln x\ln^2(1+x)}{1+x}\ dx}_{I_3}-\frac13\underbrace{\int_0^1\frac{\ln^3\left(\frac{x}{1+x}\right)}{1+x}\ dx}_{I_4}$$
$$I_1=\sum_{n=1}^\infty(-1)^{n-1}\int_0^1 x^{n-1}\ln^3x\ dx=6\sum_{n=1}^\infty\frac{(-1)^n}{n^4}=-\frac{21}4\zeta(4)$$
$$I_2=\frac14\ln^42$$
$$I_3\overset{IBP}{=}-\frac13\int_0^1\frac{\ln^3(1+x)}{x}\ dx=2\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac12\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$
Where the last result follows from using the generalization
$$\int_0^1\frac{\ln^n(1+x)}{x}\ dx=\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)$$
For $I_4$ , let $\frac{x}{1+x}\to x$
$$I_4=\int_0^{1/2}\frac{\ln^3x}{1-x}\ dx=-6\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{21}4\ln2\zeta(3)+\frac32\ln^22\zeta(2)-\frac1{12}\ln^42$$
which follows from using the generalization 
$$\int_0^{1/2}\frac{\ln^n x}{1-x}\ dx=-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}(-1)^k k!\operatorname{Li}_{k+1}\left(\frac12\right)$$
which can be found on the same link above ( check $(3)$).
Combine these results we get
$$I=4\operatorname{Li_4}\left(\frac12\right)-\frac{15}4\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac{1}{6}\ln^42$$
Giving us 

$$\int_0^1\frac{\ln x\ln^2(1+x)}{x}\ dx=-4\operatorname{Li_4}\left(\frac12\right)+\frac{15}4\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac{1}{6}\ln^42$$

A: Here is a magical way of extracting the value of the desired integral https://math.stackexchange.com/q/3531878.
A: $$I=\int_{0}^{1}\frac{\log(x)\log^2(1+x)}{x}\,dx = -\int_{0}^{+\infty} t \log(1+e^{-t})\,dt$$
is an integral that already appeared on MSE. It can be tackled in many ways, for instance by expanding $\log^2(1+x)$ as its Taylor series:
$$ \log^2(1+x) = 2\sum_{n\geq 1}\frac{(-1)^{n+1} H_n}{n+1} x^{n+1} $$
from which:
$$ I = 2\sum_{n\geq 1}\frac{(-1)^{n} H_n}{(n+1)^3} $$
follows. Evaluation of such series is a fashion topic here on MSE: Shobhit mentioned the last series (Euler sum) in his answer to a related question.
A: Performing integration by parts by taking $u=\ln^2(1+x)$ and $\mathrm dv=\dfrac{\ln x}{x}\ \mathrm dx$, then
\begin{align}
I&=\int_0^1\frac{\ln x\ln^2(1+x)}{x}\ \mathrm dx\\
&=\ln x\ln^2(1+x)\Bigg|_0^1-\int_0^1\frac{\ln^2{x}\ln(1+x)}{1+x}\,\mathrm dx\\
&=-\int_0^1\sum_{k=1}^\infty (-1)^{k-1}H_{k}\,x^k\ln^2x\,\mathrm dx\tag{1}\\
&=-\sum_{k=1}^\infty (-1)^{k-1}H_{k}\int_0^1x^k\ln^2x\,\mathrm dx\\
&=-2\sum_{k=1}^\infty (-1)^{k-1}\frac{H_{k}}{(k+1)^3}\tag{2}\\
&=-2\sum_{k=1}^\infty (-1)^{k-1}\left[\frac{H_{k+1}}{(k+1)^3}-\frac{1}{(k+1)^4}\right]\tag{3}\\
&=2\sum_{k=1}^\infty (-1)^{k-1}\left[\frac{H_{k}}{k^3}-\frac{1}{k^4}\right]\\
&=\frac{11\pi^4}{180}-4\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7\zeta(3)\ln 2}{2}+\frac{\pi^2\ln^2 2}{6}-\frac{\ln^4 2}{6}-2\eta(4)\tag{4}\\
&=\frac{11\pi^4}{180}-4\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7\zeta(3)\ln 2}{2}+\frac{\pi^2\ln^2 2}{6}-\frac{\ln^4 2}{6}-\frac{7\zeta(4)}{4}\tag{5}\\
&=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{\pi^4}{24}-4\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7\zeta(3)\ln 2}{2}+\frac{\pi^2\ln^2 2}{6}-\frac{\ln^4 2}{6}}}
\end{align}

Explanation :
$(1)$ Use generating function $\displaystyle\sum_{k=1}^\infty (-1)^{k-1}H_{k}\,x^k=\frac{\ln(1+x)}{1+x}$
$(2)$ Use formula $\displaystyle\int_0^1 x^k \ln^n x\ \mathrm dx=\frac{(-1)^n n!}{(k+1)^{n+1}}\quad,\  n\in\mathbb{Z}_{n\ge0}$
$(3)$ Use property $\displaystyle H_{k}=H_{k+1}-\frac{1}{k+1}$
$(4)$ Use the result $\displaystyle \sum_{k=1}^\infty (-1)^{k-1} \frac{H_k}{k^3} = \frac{11\pi^4}{360}-2\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7\zeta(3)\ln 2}{4}+\frac{\pi^2\ln^2 2}{12}-\frac{\ln^4 2}{12}$
$(5)$ Use property of Dirichlet eta function $\displaystyle \eta(s)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^s}=\left(1-2^{1-s}\right)\zeta(s)$
A: Note
\begin{align}
\int_{0}^{1}\frac{\ln x\ln^2(1+x)}{x}\>{dx} 
\overset{ibp} =-\int_0^1 \frac{\ln^2x \ln (1+x)}{1+x}{dx } \overset{ x\to \frac{x}{1-x} } = J_1 -J_2 +  -\frac14\ln^4 2\tag1
\end{align}
where
\begin{align}
J_1=& \int_0^{1/2} \frac{\ln^2x\ln(1-x)}{1-x} dx\\
= &\int_0^{1} \frac{\ln^2x\ln(1-x)}{1-x} dx - J_1- \frac12\ln^42\\
 =&\>\frac12\left(\int_0^1 \frac{\ln^2x}{1-x}\int_0^1 \frac {-x}{1-x t}dt\>dx - \frac12\ln^42\right)\\
=&\>\frac12\int_0^1 \frac{1}{1-t}\left(\int_0^1 \frac {\ln^2x}{1-tx}dx - 2Li_3(1)\right) dt - \frac14\ln^42\\
=&\>\int_0^1\frac{Li_3(t)}{t}dt
 + \int_0^1\frac{Li_3(t)-Li_3(1)}{1-t}\>\overset{ibp}{dt} - \frac14\ln^42\\
= &\>Li_4(1) - \frac12Li_2^2(1) - \frac14\ln^42 = -\frac{\pi^4}{360} - \frac14\ln^42\\
\\
J_2=&
\>2 \int_0^{1/2} \frac{\ln x\ln^2(1-x)}{1-x}\>\overset{ibp}{dx}\\
 =&-\frac23\ln^42 + \frac23
\int_0^{1/2} \frac{\ln^3(1-x)}{x} \>\overset{x\to 1-x}{dx}  \\
= & -\frac23\ln^42 -4Li_4(1) -\frac23\int_0^{1/2} \frac{\ln^3x}{1-x} \overset{x\to x/2}{dx} \\
=&-4Li_4(1) +4Li_4(\frac12)+4\ln2Li_3(\frac12) +2\ln^2Li_2(\frac12)\\
= & \>4Li_4(\frac12)+\frac72\ln2\zeta(3)-\frac{2\pi^4}{45}-\frac{\pi^2}6\ln^22-\frac13\ln^42\\
\\
\end{align}
Substitute $J_1$ and $J_2$ into (1) to obtain
$$\int_{0}^{1}\frac{\ln x\ln^2(1+x)}{x}\>{dx} 
= -4Li_4(\frac12)-\frac72\ln2\zeta(3)+\frac{\pi^4}{24}+\frac{\pi^2}6\ln^22-\frac1{6}\ln^42
$$
