AIが出力した積分を解く②

$$\begin{array}{rcl}{\int }_0^1\frac{x^2{\ln }^2(x)}{(1-x^2{)}^2}dx& =& \frac{1}{4}{\int }_0^1{\ln }^2(x)(\frac{1}{(x-1{)}^2}+\frac{1}{(x+1{)}^2}+\frac{1}{x-1}-\frac{1}{x+1})dx\\ & =& \frac{1}{4}({\int }_0^1\frac{{\ln }^2(x)}{(x-1{)}^2}dx+{\int }_0^1\frac{{\ln }^2(x)}{(x+1{)}^2}dx+{\int }_0^1\frac{{\ln }^2(x)}{x-1}dx-{\int }_0^1\frac{{\ln }^2(x)}{x+1}dx)\end{array}$$
$$\begin{array}{rcl}{\int }_0^1\frac{{\ln }^2(x)}{(x-1{)}^2}dx& =& {\int }_0^1{\ln }^2(x)(1+x+x^2+\cdots )(1+x+x^2+\cdots )dx\\ & =& {\int }_0^1{\ln }^2(x)\sum _{n=0}^{\mathrm{\infty }}(n+1)x^{n}dx\\ & =& \sum _{n=0}^{\mathrm{\infty }}{\int }_0^1(n+1){\ln }^2(x)x^{n}dx\\ & =& \sum _{n=0}^{\mathrm{\infty }}\frac{2(n+1)}{(n+1{)}^3}=2\zeta (2)=\frac{{\pi }^2}{3}\end{array}$$

$$\begin{array}{rcl}{\int }_0^1\frac{{\ln }^2(x)}{(x+1{)}^2}dx& =& {\int }_0^1{\ln }^2(x)\sum _{n=0}^{\mathrm{\infty }}(n+1)(-1{)}^n{x}^{n}dx\\ & =& \sum _{n=0}^{\mathrm{\infty }}\frac{2(n+1)(-1{)}^n}{(n+1{)}^3}\\ & =& 2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+1{)}^2}-4\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n{)}^2}=2\zeta (2)-\zeta (2)=\frac{{\pi }^2}{6}\end{array}$$

$$\begin{array}{rcl}{\int }_0^1\frac{{\ln }^2(x)}{x-1}dx& =& -{\int }_0^1{\ln }^2(x)\sum _{n=0}^{\mathrm{\infty }}{x}^{n}dx\\ & =& -\sum _{n=0}^{\mathrm{\infty }}\frac{2}{(n+1{)}^3}=-2\zeta (3)\end{array}$$

$$\begin{array}{rcl}{\int }_0^1\frac{{\ln }^2(x)}{x+1}dx& =& {\int }_0^1{\ln }^2(x)\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n{x}^{n}dx\\ & =& \sum _{n=0}^{\mathrm{\infty }}\frac{2(-1{)}^n}{(n+1{)}^3}\\ & =& 2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+1{)}^3}-4\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n{)}^3}=2\zeta (3)-\frac{1}{2}\zeta (3)=\frac{3}{2}\zeta (3)\end{array}$$

$$\therefore {\int }_0^1\frac{x^2{\ln }^2(x)}{(1-x^2{)}^2}dx=\frac{1}{4}(\frac{{\pi }^2}{3}+\frac{{\pi }^2}{6}-2\zeta (3)-\frac{3}{2}\zeta (3))=\frac{1}{8}({\pi }^2-7\zeta (3))$$