Proving $\int_0^1 \frac{(\ln(x))^5}{1+x} \mathrm{d}x = -\frac{31\pi^6}{252}$

HINT:

Enforce the substitution $x\to e^{-x}$, expand the resulting denominator in a geometric series of $\sum_{n=0}^{\infty}(-1)^ne^{-nx}$, interchange the sum and integral, carryout the integral by either successive IBP or differentiating under the integral, and evaluate the resulting series representation of $\zeta(6)$.

Alternatively, note that

$$\int_0^1 \frac{\log^5(x)}{1+x}\,dx=\left. \left(\frac{d^5}{da^5}\int_0^1\frac{x^a}{1+x}\,dx\right)\right|_{a=0}$$

Let $\Re(s)>0$ and $$J(s)=\int_0^1\frac{\ln^sx}{1+x}dx.$$ Letting $x=e^{-u}$, $$\begin{align} J(s)&=(-1)^s\int_0^\infty\frac{u^se^{-u}}{1+e^{-u}}du\\ &=(-1)^s\int_0^\infty u^s\sum_{n\ge1}(-1)^ne^{-nu}du\\ &=(-1)^s\sum_{n\ge1}(-1)^n\int_0^\infty u^se^{-nu}du\\ &=(-1)^s\sum_{n\ge1}\frac{(-1)^n}{n^{s+1}}\int_0^\infty u^se^{-u}du\\ &=(-1)^s\Gamma(s+1)\sum_{n\ge1}\frac{(-1)^n}{n^{s+1}}\\ &=(-1)^s(2^{-s}-1)\Gamma(s+1)\zeta(s+1). \end{align}$$