Who's up to the challenge? 19

Using expansion for $ln(1-x^2)$

$\displaystyle -I = \int_{0}^{1} (lnx)^3 \left( \sum_{k=1}^{ \infty } \dfrac{x^{2k}}{k} \right) dx$

Interchanging sings of summation and integration

$\displaystyle -I= \sum_{k=1}^{ \infty } \dfrac{1}{k} \int_{0}^{1} (lnx)^3 x^{2k} dx$

Integrating by parts 3 times we get

$\dfrac{-I}{6} = \sum_{k=1}^{ \infty } \dfrac{1}{k(2k+1)^4}$

Using partial fractions we get

$\displaystyle \dfrac{-I}{6} = \sum_{k=1}^{ \infty } \left( \dfrac{-2}{(2k+1)^4} +\dfrac{-2}{(2k+1)^3} + \dfrac{-2}{(2k+1)^2} +\dfrac{2}{2k(2k+1)} \right)$

Now sums are easy. solving them we get

$\displaystyle I = \dfrac{-21}{2} \zeta{(3)} - \dfrac{3 \pi^2}{2} + 48 - \dfrac{ \pi^4}{8} -12 \ln(2)$