10. $[8]$ Evaluate the integral $\int_{0}^{1} \ln x \ln (1-x) d x$.
Answer: $2-\frac{\pi^{2}}{6}$ We have the MacLaurin expansion $\ln (1-x)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\cdots$. So
$$
\int_{0}^{1} \ln x \ln (1-x) d x=-\int_{0}^{1} \ln x \sum_{n=1}^{\infty} \frac{x^{n}}{n} d x=-\sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} x^{n} \ln x d x .
$$

Using integration by parts, we get
$$
\int_{0}^{1} x^{n} \ln x d x=\left.\frac{x^{n+1} \ln x}{n+1}\right|_{0} ^{1}-\int_{0}^{1} \frac{x^{n}}{n+1} d x=-\frac{1}{(n+1)^{2}} .
$$
(We used the fact that $\lim _{x \rightarrow 0} x^{n} \ln x=0$ for $n>0$, which can be proven using l'Hôpital's rule.) Therefore, the original integral equals to
$$
\sum_{n=1}^{\infty} \frac{1}{n(n+1)^{2}}=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}-\frac{1}{(n+1)^{2}}\right) .
$$

Telescoping the sum and using the well-known identity $\sum_{n=0}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$, we see that the above sum is equal to $2-\frac{\pi^{2}}{6}$.