Another Headache for you

If $\displaystyle J(a)=\int_{0}^{1} x^a dx=\frac{1}{a+1}$ , then $\displaystyle J^{(b)}(a) = \int_{0}^{1}x^a (\ln x)^b dx= \frac{(-1)^b b!}{(a+1)^{b+1}}$

$\displaystyle S=\int_{0}^{1} \frac{(\ln x)^4}{(1-x)^4} dx = \sum_{m=1}^{\infty} \binom{m+2}{3} \int_{0}^{1} x^m \ln^4 x dx = \sum_{m=1}^{\infty} \binom{m+2}{3} J^{(4)}(m)$

So , $\displaystyle S=\sum_{m=1}^{\infty} \frac{(m+2)!}{3!(m-1)!} \frac{(-1)^4 4!}{m^5}$

$\displaystyle S = \sum_{m=1}^{\infty} 4 \frac{(m+1)(m+2)}{m^4} = \sum_{m=1}^{\infty} \frac{4}{m^2}+\frac{12}{m^3}+\frac{8}{m^4} = 4\zeta(2)+12\zeta(3)+8\zeta(4)$

Pretty simple.Just expand (1-x)^(-4) in series form then use gamma function definition