Proving $\int_0^1 \frac{(\ln(x))^5}{1+x} \mathrm{d}x = -\frac{31\pi^6}{252}$ I would like to show the following identity:

$$\boxed{
I := 
\int_0^1 \dfrac{(\ln(x))^5}{1+x} \mathrm{d}x =
-\dfrac{31\pi^6}{252}
}$$


Here is what I tried.
The change of variables $u=1/x$ yields
$$I=
\int_1^{\infty} \dfrac{(\ln(x))^5}{1+1/u} \dfrac{1}{u^2} \mathrm{d}u =
\int_1^{\infty} \dfrac{(\ln(x))^5}{u^2+u} \mathrm{d}u$$
Then $z=u-1$ gives
$$I = \int_{0}^{\infty} \dfrac{(\ln(z+1))^5}{z^2+3z+2} \mathrm{d}z $$
with $z^2+3z+2=(z+1)(z+2)$.
I wanted to use contour integration like here, but I was not sure how to proceed in this case. Anyway, the computations of the residues (of which "well-chosen" function? Maybe something like this?) seem to be difficult.
I believe that we can generalize to $\frac{(\ln(x))^n}{1+x}$, or maybe even more (e.g. $\frac{(\ln(x))^n}{ax^2+bx+c}$).
Related computations are: (1), (2), (3), (4).
Thank you for your detailed help.
 A: 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
$$\int_0^1 \frac{\log^5(x)}{1+x}\,dx=\left. \left(\frac{d^5}{da^5}\int_0^1\frac{x^a}{1+x}\,dx\right)\right|_{a=0}$$
A: Let $\Re(s)>0$ and
$$J(s)=\int_0^1\frac{\ln^sx}{1+x}dx.$$
Letting $x=e^{-u}$,
$$\begin{align}
J(s)&=(-1)^s\int_0^\infty\frac{u^se^{-u}}{1+e^{-u}}du\\
&=(-1)^s\int_0^\infty u^s\sum_{n\ge1}(-1)^ne^{-nu}du\\
&=(-1)^s\sum_{n\ge1}(-1)^n\int_0^\infty u^se^{-nu}du\\
&=(-1)^s\sum_{n\ge1}\frac{(-1)^n}{n^{s+1}}\int_0^\infty u^se^{-u}du\\
&=(-1)^s\Gamma(s+1)\sum_{n\ge1}\frac{(-1)^n}{n^{s+1}}\\
&=(-1)^s(2^{-s}-1)\Gamma(s+1)\zeta(s+1).
\end{align}$$
A: \begin{align}J&=\int_0^1 \frac{(\ln(x))^5}{1+x} \mathrm{d}x \\
&=\int_0^1 \frac{(\ln(x))^5}{1-x} \mathrm{d}x-\int_0^1 \frac{2u(\ln(u))^5}{1-u^2} \mathrm{d}u\\
&\overset{x=u^2}=\int_0^1 \frac{(\ln(x))^5}{1-x} \mathrm{d}x-\frac{1}{2^5}\int_0^1 \frac{(\ln(x))^5}{1-x} \mathrm{d}x\\
&=\frac{31}{32}\int_0^1 \frac{(\ln(x))^5}{1-x} \mathrm{d}x\\
&=-\frac{31}{32}\times \frac{5!}{945}\times \zeta(6)\\
&=\boxed{-\frac{31\pi^6}{252}}
\end{align}
I assume: $\displaystyle \int_0^1 \frac{\ln^5 x}{1-x}dx=-5!\zeta(6)$ and $\displaystyle\zeta(6)=\frac{\pi^6}{945}$
A: Use a geometric series.
$$\frac{1}{1+x}=\sum_{k=0}^\infty(-1)^kx^k$$
So
$$\mathcal{I}_n=\int_0^1\frac{\ln(x)^n}{1+x}\mathrm{d}x=\sum_{m=0}^\infty\left[(-1)^mI_{m,n}\right]$$
Where
$$I_{m,n}=\int_0^1 x^m\ln(x)^n\mathrm{d}x$$
Now use a substitution $x=e^{-t}$ which changes our region of integration to $(\infty,0]$.
$$I_{m,n}=\int_\infty^0(e^{-t})^m(-t)^n(-e^{-t})\mathrm{d}t$$
$$=(-1)^n\int_0^\infty t^ne^{-t(m+1)}\mathrm{d}t$$
Now use another substitution $z=t(m+1)$:
$$I_{m,n}=(-1)^n\int_0^\infty\left(\frac{z}{m+1}\right)^ne^{-z}\frac{\mathrm{d}z}{m+1}$$
$$=\frac{(-1)^n ~n!}{(m+1)^{n+1}}$$
So then
$$\mathcal{I}_n=(-1)^nn!\sum_{m=0}^\infty\frac{(-1)^m}{(m+1)^{n+1}}$$
$$=(-1)^nn!\sum_{m=1}^\infty\frac{(-1)^{m-1}}{m^{n+1}}$$
$$=(-1)^n~n!~\eta(n+1)$$
Using the Dirichlet eta function. In the case of $n=5$ we get
$$\mathcal{I}_5=-5!\cdot\eta(6)=\frac{-31\pi^6}{252}$$
Which is confirmed numerically by Mathematica:

