Evaluate integral: $\int_0^1 \frac{\log^3(1-x)}{x}dx$ I need your help to evaluate the following integral:
$$I=\int_0^1 \frac{\log^3(1-x)}{x}dx$$
Using the fact that for $|x|<1$ I get
$$\log(1-x)=-\sum_{n=1}^{\infty}\frac{x^k}{k}$$
One can rewrite $I$ as
$$I=-\sum_{k=0}^{\infty}\frac{1}{k+1}\left(\int_0^1 x^k\log^2(1-x)dx\right)$$
I tried to rewrite $\log(1-x)$ as a sum but I get some "monstrous" summation to calculate. 
 A: $$
\begin{align}
\int_0^1\frac{\log^3(1-x)}{x}\,\mathrm{d}x
&=-\int_0^\infty\frac{u^3}{1-e^{-u}}e^{-u}\,\mathrm{d}u\tag{1}\\
&=-\int_0^\infty u^3\sum_{k=1}^\infty e^{-ku}\,\mathrm{d}u\tag{2}\\
&=-\sum_{k=1}^\infty\frac1{k^4}\int_0^\infty u^3e^{-u}\,\mathrm{d}u\tag{3}\\[6pt]
&=-\zeta(4)\,\Gamma(4)\tag{4}\\[9pt]
&=-\frac{\pi^4}{90}\cdot6\tag{5}\\[6pt]
&=-\frac{\pi^4}{15}\tag{6}
\end{align}
$$
Explanation:
$(1)$: $x=1-e^{-u}$
$(2)$: series $\frac{x}{1-x}=\sum\limits_{k=1}^\infty x^k$
$(3)$: substitute $u\mapsto\frac uk$
$(4)$: use a couple of special functions
$(5)$: evaluate special functions (see this answer for $\zeta(4)$)
$(6)$: simplification

Addendum
Note that just as
$$
\int_0^\infty\frac{x^{a-1}}{e^x}\,\mathrm{d}x=\Gamma(a)
$$
we have
$$
\int_0^\infty\frac{x^{a-1}}{e^x-1}\,\mathrm{d}x=\Gamma(a)\zeta(a)
$$
A: Since robjohn already provided a detailed solution, I will just outline a non-trivial consequence of the substitution $x\mapsto(1-x)$. Since $\int_{0}^{1} x^m\log^3(x)\,dx=-\frac{6}{(m+1)^3}$ for any $m\in\mathbb{N}$, the original integral equals $-6\,\zeta(4)=-\frac{\pi^4}{15}$. On the other hand:
$$ \log(1-x) = \sum_{n\geq 1}\frac{x^n}{n},\qquad \log^2(1-x)=\sum_{n\geq 1}\frac{2 H_{n-1}}{n}x^{n} \tag{1} $$
lead, by summation by parts, to
$$ \log^3(1-x) = -3\sum_{n\geq 1}\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n}x^n \tag{2}$$
so:
$$ \int_{0}^{1}\frac{\log^3(1-x)}{x}\,dx = -3\left[\sum_{n\geq 1}\frac{H_{n-1}^2}{n^2}-\sum_{n\geq 1}\frac{H_{n-1}^{(2)}}{n^2}\right]\tag{3} $$
and since
$$ \sum_{n\geq 1}\frac{H_{n-1}^{(2)}}{n^2} = \sum_{1\leq m < n}\frac{1}{m^2 n^2}=\frac{\zeta(2)^2-\zeta(4)}{2} = \frac{\pi^4}{120}\tag{4} $$
surprising consequences of the substitution $x\mapsto(1-x)$ are:
$$ \sum_{n\geq 1}\frac{H_{n-1}^2}{n^2}=\frac{11\pi^4}{360},\qquad \sum_{n\geq 1}\frac{H_n^2}{n^2}=\frac{17\pi^4}{360}.\tag{5}$$
As a further reference, please see pages 91-92 here.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
I & \equiv \int_{0}^{1}{\ln^{3}\pars{1 - x} \over x}\,\dd x
\,\,\,\stackrel{x\ \mapsto\ 1 - x}{=}\,\,\,
\int_{0}^{1}{\ln^{3}\pars{x} \over 1 - x}\,\dd x
\\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\,
\int_{0}^{1}\ln\pars{1 - x}\bracks{3\ln^{2}\pars{x}
\,{1 \over x}}\dd x =
-3\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}
\,\dd x
\\[5mm] &\ \pars{~\substack{\ds{\mrm{Li}_{s}:\ PolyLogarithm\ Function\,,\quad \mrm{Li}_{s}\pars{0} = 0}\\[2mm]
\ds{\mrm{Li}_{s + 1}'\pars{z} = {\mrm{Li}_{s}\pars{z} \over z}\,,\quad\mrm{Li}_{1}\pars{z} = -\ln\pars{1 - z}}}~}
\\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\,
3\int_{0}^{1}\mrm{Li}_{2}\pars{x}\bracks{2\ln\pars{x}\,{1 \over x}}\,\dd x =
6\int_{0}^{1}\mrm{Li}_{3}'\pars{x}\ln\pars{x}\,\dd x
\\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\,
-6\int_{0}^{1}\mrm{Li}_{3}\pars{x}\,{1 \over x}\,\dd x
=
-6\int_{0}^{1}\mrm{Li}_{4}'\pars{x}\,\dd x = -6\,\mrm{Li}_{4}\pars{1}
\\[5mm] & =
-6\ \underbrace{\zeta\pars{4}}_{\ds{\pi^{4} \over 90}}
\qquad\qquad\qquad\pars{~\substack{\ds{\zeta:\ Riemann\ Zeta\ Function}\\[2mm]
\ds{\mrm{Li}_{s}\pars{1} = \zeta\pars{s}}}~}
\\[5mm] & =
\bbx{-\,{\pi^{4} \over 15}} \\ &
\end{align}
