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.

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.
