\int _0^1\ln  x \ln (1-x)dx=-\int _0^1\ln  x \sum _{n=1}^{\infty } \frac{x^n}{n}dx=-\int_0^1 \left(\sum _{n=1}^{\infty }  \frac{x^n}{n}\ln  x\right) \, dx=-\sum _{n=1}^{\infty } \int _0^1\frac{x^n}{n}\ln  xdx