Show that $\int_0^1 \frac{\ln^3(x)}{1+x} dx = \sum_{k=0}^{\infty}(-1)^{k+1} \frac6{(k+1)^4}$ Show that;

$$ \int_0^1 \frac{\ln^3(x)}{1+x} dx = \sum_{k=0}^{\infty}(-1)^{k+1} \frac6{(k+1)^4}$$ 

I arrived to the fact that 
$$ \int_0^1 \frac{\ln^3(x)}{1+x} dx = \sum_{k=0}^{\infty}(-1)^k\int_0^1 x^k \ln^3(x)dx$$
But I am unable to continue further.
 A: An alternative approach is to use differentiation under the integral sign. For any $\alpha > 0$ we have
$$ \int_{0}^{1}\frac{x^{\alpha}}{1+x}\,dx = \sum_{n\geq 0}(-1)^n\int_{0}^{1}x^{\alpha+n}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{n+\alpha+1}\tag{1}$$
hence by differentiating three times wrt $\alpha$ both sides of $(1)$ we get:
$$ \int_{0}^{1}\frac{x^{\alpha}\log^3(x)}{1+x}\,dx = \sum_{n\geq 0}\frac{6(-1)^{n+1}}{(n+\alpha+1)^4}\tag{2}$$
and by considering the limit s $\alpha\to 0^+$:
$$ \int_{0}^{1}\frac{\log^3(x)}{1+x}\,dx = 6\sum_{n\geq 1}\frac{(-1)^n}{n^4} = -\frac{21}{4}\zeta(4)=\color{red}{-\frac{7\pi^4}{120}}\tag{3} $$
follows.
A: Just integrate by parts:
\begin{align*}
\int_0^1 x^k \ln^3(x) \, dx &= \frac{x^{k + 1}}{k + 1} \ln^3 x \big|_0^1 - \int_0^1 \frac{x^{k + 1}}{k + 1} 3 \ln^2(x) \frac 1 x \, dx \\
&= -\frac{3}{k + 1} \int_0^1 x^k \ln^2(x) \, dx
\end{align*}
The next application reverses the sign, picks up factor $2/(k + 1)$, and drops the power on the log by $1$. Continue until the log disappears.
A: The substitution $u=-\ln x$ rewrites the integral as $$-\int_0^\infty \frac{u^3e^{-u}du}{1+e^{-u}}=\sum_{k=0}^\infty\left(-1\right)^{k+1}\int_0^\infty u^3e^{-\left(k+1\right)u}du.$$ The substitution $v=\left(k+1\right)u$ in $\int_0^\infty v^3 e^{-v}dv=3!=6$ obtains $$\int_0^\infty u^3e^{-\left(k+1\right)u}du=\frac{6}{\left(k+1\right)^4},\,\int_0^1\frac{\ln^3x}{1+x}dx=6\sum_{k=0}^\infty\frac{\left(-1\right)^{k+1}}{\left(k+1\right)^4}.$$
A: We begin with the substitution $$x=\mathrm{e}^{-y}$$ so our integral becomes
\begin{equation}
-\int\limits_{0}^{\infty} \frac{y^{3}}{1+\mathrm{e}^{\,y}} \mathrm{d} y
\end{equation}
This integral is a Mellin transform of the function
\begin{equation}
f(y) = \frac{1}{1+\mathrm{e}^{\,y}}
\end{equation}
where the Mellin transform is defined as
\begin{equation}
\mathcal{M}[f(x](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x
\end{equation}
From Tables of Integral Transforms (Bateman Manuscript) Volume 1, we have
\begin{equation}
\mathcal{M}[(\mathrm{e}^{\alpha x} + 1)^{-1}](s) = \frac{1}{\alpha^{s}} \Gamma(s) (1-2^{1-s}) \zeta(s)
\label{eq:1}
\tag{1}
\end{equation}
With $$s = 4 \quad \mathrm{and} \quad \alpha = 1$$ we have
\begin{equation}
-\int\limits_{0}^{\infty} \frac{y^{3}}{1+\mathrm{e}^{y}} \mathrm{d} y = -\Gamma(4) (1-2^{1-4}) \zeta(4) = -\frac{7\pi^{4}}{120}
\end{equation}
Thus
\begin{equation}
\int\limits_{0}^{1} \frac{\mathrm{ln}^{3}(x)}{1+x} \mathrm{d} x = -\frac{7\pi^{4}}{120}
\end{equation}
Addendum
We can rewrite \eqref{eq:1} as
\begin{align}
\mathcal{M}[(\mathrm{e}^{\alpha x} + 1)^{-1}](s) & = \frac{1}{\alpha^{s}} \Gamma(s) (1-2^{1-s}) \zeta(s) \\
& = \frac{1}{\alpha^{s}} \Gamma(s) \eta(s)
\end{align}
where
\begin{equation}
\eta(s) = (1-2^{1-s}) \zeta(s) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}
\end{equation}
is the Dirichlet eta function, also known as the alternating Riemann zeta function.
Note that for Wolfram Alpha, $\eta(s)$ defaults to the Dedekind eta function. To obtain the Dirichlet eta function type "dirichlet eta(s)".
A: Substitute $x=e^{-u}$:
$$
\begin{align}
\int_0^1\frac{\log^3(x)}{1+x}\,\mathrm{d}x
&=-\int_0^\infty\frac{u^3}{1+e^{-u}}e^{-u}\,\mathrm{d}u\\
&=\int_0^\infty\sum_{k=1}^\infty(-1)^ku^3e^{-ku}\,\mathrm{d}u\\
&=\int_0^\infty u^3e^{-u}\sum_{k=1}^\infty\frac{(-1)^k}{k^4}\,\mathrm{d}u\\
&=3!\sum_{k=1}^\infty\frac{(-1)^k}{k^4}\\
\end{align}
$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x} & =
\lim_{\mu \to 0}\,\partiald[3]{}{\mu}\int_{0}^{1}{x^{\mu} \over 1 + x}\,\dd x =
\lim_{\mu \to 0}\,\partiald[3]{}{\mu}
\int_{0}^{1}{x^{\mu} - x^{\mu + 1} \over 1 - x^{2}}\,\dd x
\\[5mm] & \stackrel{x^{2}\ \mapsto\ x}{=}\
\half\,\lim_{\mu \to 0}\,\partiald[3]{}{\mu}\bracks{%
\int_{0}^{1}{x^{\mu/2 - 1/2}\,\,\, -\,\,\, x^{\mu/2} \over 1 - x}\,\dd x}
\\[5mm] & =
\half\,\lim_{\mu \to 0}\,\partiald[3]{}{\mu}\bracks{%
\int_{0}^{1}{1 - x^{\mu/2} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{\mu/2 - 1/2} \over 1 - x}\,\dd x}
\\[5mm] & =
\half\,\lim_{\mu \to 0}\,\partiald[3]{}{\mu}
\bracks{\Psi\pars{\mu + 2 \over 2} - \Psi\pars{\mu + 1 \over 2}}\tag{1}
\end{align}
where $\ds{\Psi}$ is the Digamma Function. In $\ds{\pars{1}}$ we used the
identity:
$\ds{\Psi\pars{\xi} = -\gamma + \int_{0}^{1}{1 - t^{\xi - 1} \over 1 - t}
\,\dd \xi}$ with $\ds{\Re\pars{\xi} > 0}$.

\begin{align}
\color{#f00}{\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x} & =
{1 \over 16}\bracks{\Psi'''\pars{1} - \Psi'''\pars{\half}} =
\color{#f00}{-\,{7 \over 120}\,\pi^{4}} \approx -5.6822
\end{align}


Some values of $\ds{\Psi'''}$ are well known:
  $\ds{\Psi'''\pars{1} = 3! \times \zeta\pars{4} = {1 \over 15}\,\pi^{4}\,,\qquad
\Psi'''\pars{\half} = 3! \times 15 \times \zeta\pars{4} = \pi^{4}}$
  with
  $\ds{\zeta\pars{4} = {1 \over 90}\,\pi^{4}}$.

