Table of Integrals, Series, and Products, 8th Edition

$\begin{array}{ll}{\displaystyle {\int }_0^1{\left(ln\frac{1}{1-x^2}\right)}^n\text{?}{x}^{2q-1}\text{?}\text{d}x=\frac{n!}{2}}\text{?}\zeta (n+1,\text{?}q+1)\hfill & [-1<q<0]\hfill \end{array}$

  BI (311)(15)a

12. 

${{\displaystyle \int }}_0^1{(lnx)}^{2n}\text{?}ln\text{?}(1-x^2)\text{?}\frac{\text{d}x}{x}=-\frac{{\pi }^{2n+2}}{2(n+1)(2n+1)}\left|B_{2n+2}\right|$

  BI (309)(5)a

13.⁶ 

$\begin{array}{ll}{\displaystyle {\int }_0^1{\left[ln\text{?}\frac{1}{x}\right]}^m\text{?}ln\text{?}(1-x^2)\text{?}\text{d}x=-}{\displaystyle \sum _{n=1}^{\infty }\frac{\mathrm{\Gamma }(m+1)}{n{(2n+1)}^{m+1}}}\hfill & [m+1>0,\text{?}n+1>0]\hfill \end{array}$

4.295

1. 

$\begin{array}{ll}{\displaystyle {\int }_0^{\infty }\frac{ln(a{x}^2+b)}{1+x^2}\text{?}\text{d}x=\pi \text{?}ln\text{?}(\sqrt{a}+\sqrt{b})}\hfill & [\mathrm{Re}\text{?}a>0,\text{?}\mathrm{Re}\text{?}b>0]\hfill \end{array}$

  ET II 218(27)

2. 

${\displaystyle {\int }_0^{1}ln\text{?}(1+x^2)\frac{\text{d}x}{x^2}=\frac{\pi }{2}-ln\text{?}2}$

  GW (325)(2g)

3. 

${\displaystyle {\int }_0^{\infty }ln\text{?}(1+x^2)\frac{\text{d}x}{x^2}=}$