$1.\,\bold{Contents}$
:
$$\begin {aligned}
\int _{0}^{1}\cos (\pi x)\ln \frac {\Gamma \left (\frac {x}2\right )}{\Gamma \left (\frac {x+1}2\right )}dx
&=\frac{1}2
\end{aligned}$$
:
$$\begin {aligned}
\int _{0}^{\infty }\frac {\ln (x)\ln (1+x)}{(1+x)(1+x^{2})}dx
&=-\frac {35}{64}\zeta (3)+\frac {\pi ^{3}}{32}-\frac {\pi ^{2}\ln 2}{32}
\\
\int _{0}^{1}\frac {(1-x)\ln (x)\ln (1-x)}{(1+x)(1+x^{2})}dx
&=\frac {7}{64}\zeta (3)-\frac {\pi ^{2}\ln 2}{32}
\end {aligned}$$
:
$$\begin {aligned}
\int _{0}^{\infty }\frac {\ln ^{2}x}{(1+x)^2}dx
&=\frac {\pi ^{2}}3
\\
\int _{0}^{\infty }\frac {\ln ^{2}x}{(1+x^{2})^{2}}dx
&=\frac {\pi ^{3}}{16}
\end {aligned}$$
:
$$\lim _{r\to 0+}\left (\frac {\ln r}2+\int _{0}^{1-r}\frac {dx}{(1-x)(1+x^{2})}\right )
=\frac {\pi }8+\frac {\ln 2}4$$
:
$$\begin {aligned}
\int _{0}^{1}\frac {\ln ^{2}x}{(1-x)(1+x^{2})}dx
&=\frac {35}{32}\zeta (3)+\frac {\pi ^{3}}{32}
\end {aligned}$$
:
$$\int _{0}^{\frac {\pi }2}\ln \left (2\sin \frac {x}2\right )\ln \left (2\cos \frac {x}2\right )dx
=-\frac {\pi ^{3}}{48}$$
$$\int _{0}^{\frac {\pi }4}\ln (\sin x)\ln (\cos x)dx
=-\frac {\pi ^{3}}{96}+\frac {\pi \ln ^{2}2}4$$
:
$$\begin {aligned}
\int _{0}^{1}\frac {\arcsin \sqrt x}{1+x}dx
&=\frac {3\pi \ln 2}2-\pi \ln (1+\sqrt 2)
\end {aligned}$$
:
$$\begin {aligned}
\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {dxdydz}{\sqrt {1-x^{2}y^{2}z^{2}}}
&=\frac {\pi ^{3}}{48}+\frac {\pi \ln ^{2}2}{4}
\end {aligned}$$
:
$$\int _{0}^{1}\frac {\ln (1+x^{2})}{1+x^{2}}dx
=\frac {\pi \ln 2}2-G$$
:
$$\begin {aligned}
\int _{0}^{\frac {\pi }4}\ln ^2\left (\sin x\right )dx
&=\frac {\pi ^{3}}{192}+\frac {3\pi \ln ^{2}2}{16}+\frac {\ln (2)\beta (2)}2+\Im \operatorname{Li} _{3}(1+i)\\
\int _{0}^{\frac {\pi }4}\ln ^{2}(\cos x)dx
&=\frac {7\pi ^{3}}{192}+\frac {5\pi \ln ^{2}2}{16}-\frac {\ln (2)\beta (2)}2-\Im \operatorname{Li} _{3}(1+i)
\end {aligned}$$
:
$$\int _{0}^{2}\frac {dx}{\sqrt {1+x^{3}}}
=\frac {\Gamma \left (\frac {1}3\right )^3}{2^{\frac {4}3}\sqrt 3\pi }$$
:
$$\int _{0}^{\infty }\arctan (\cosh x)\ln (\tanh x)dx
=\pi \left (\operatorname{Li} _{2}\left (1-\sqrt 2\right )-\frac {1}4\operatorname {Li} _2 \left (-3+2\sqrt 2\right ) \right )$$
:
$$\int _{0}^{\frac {\pi }4}\frac {x\ln \tan x}{\sin 2x}dx
=-\frac {\pi ^{3}}{64}$$
:
$$\int _{0}^{\infty }\ln \left (\frac {1+x^{2}+x^{4}+x^{6}}{1+x^{6}}\right )dx
=\pi \left (\sqrt 2-1\right )$$
:
$$\begin {aligned}
\int _{0}^{\infty }\frac {\lfloor x\rfloor}{1+e^{2\pi x}}dx
&=\frac {1}{24}-\frac {3\ln 2}{16\pi }
\end {aligned}$$
:
$$\begin {aligned}
\int _{0}^{\frac {\pi }2}\arctan \left (2\tan ^{2}x\right )dx
&=\pi \arctan \frac {1}2
\end {aligned}$$
:
$$\begin {aligned}
\int _{0}^{\infty }\arctan \frac {\sinh x}{1+\cosh 2x}dx
&= \frac {2}3\beta (2)
\end {aligned}$$