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.

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)".
