How do I start this integral problem: $\int_0^1 \frac{\ln^3 u}{1-u} du = -\frac{\pi^4}{15} $?

$$ \begin{align*} \int_0^1 \frac{\ln^3(x)}{1-x}dx &= \int_0^1\ln^3(x)\sum_{n=0}^\infty x^n \; dx\\ &= \sum_{n=0}^\infty \int_0^1 x^n \ln^3(x) \; dx \\ &= -6\sum_{n=0}^\infty \frac{1}{(n+1)^4} \\ &= -6 \zeta(4) \\ &= -6 \times \frac{\pi^4}{90} \\ &= -\frac{\pi^4}{15} \end{align*} $$

Using $x=-\ln u$ gives \begin{align} \int_0^1\frac{(\ln u)^r}{1-u}\,du &=(-1)^r\int_0^\infty\frac{x^r e^{-x}}{1-e^{-x}}\,dx\\ &=(-1)^r\sum_{n=1}^\infty\int_0^\infty x^r e^{-nx}\,dx\\ &=(-1)^r\sum_{n=1}^\infty\frac 1{n^{r+1}}\int_0^\infty x^r e^{-x}\,dx\\ &=(-1)^r\zeta(r+1)\Gamma(r+1). \end{align} When $r$ is a nonnegative integer, $\Gamma(r+1)=r!$. So when $r=3$ this is $-6\zeta(4)=-\pi^4/15$.

$\displaystyle \int_0^1 \frac{\ln^3 u}{1-u} \, du = \int_0^1 \ln^3{u} \sum_{k \ge 0} u^k \,{du}= \sum_{k \ge 0} \int_0^1 u^k \ln^3{u} \, {du}$

Let $\displaystyle f(k) = \int_0^1 u^k \, {du} = \frac{1}{(1+k)}$. Differentiating w.r.t. $k$

We have $\displaystyle f^{(3)}(k) = \int_0^1 u^k \ln^3(u)\,{du} = \frac{-6}{(1+k)^4} $.

Hence $\displaystyle I = -6 \sum_{k\ge 0} \frac{1}{(1+k)^4} = - 6\zeta(4) = -\frac{\pi^4}{15}. $