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$\begin{array}{ll}{\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 \left(n+1,\text{?}q+1\right)\hfill & \left[-1

BI (311)(15)a

12.

${\int }_{0}^{1}{\left(lnx\right)}^{2n}\text{?}ln\text{?}\left(1-{x}^{2}\right)\text{?}\frac{\text{d}x}{x}=-\frac{{\pi }^{2n+2}}{2\left(n+1\right)\left(2n+1\right)}|{B}_{2n+2}|$

BI (309)(5)a

13.6

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

4.295

1.

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

ET II 218(27)

2.

${\int }_{0}^{1}ln\text{?}\left(1+{x}^{2}\right)\frac{\text{d}x}{{x}^{2}}=\frac{\pi }{2}-ln\text{?}2$

GW (325)(2g)

3.

${\int }_{0}^{\infty }ln\text{?}\left(1+{x}^{2}\right)\frac{\text{d}x}{{x}^{2}}=$

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