$$\begin {aligned} \int _{0}^{1}\cos (\pi x)\ln \frac {\Gamma \left (\frac {x}2\right )}{\Gamma \left (\frac {x+1}2\right )}dx &=\frac{1}2 \end{aligned}$$

$$\begin {aligned} 　\int _{0}^{\infty }\frac {\ln (x)\ln (1+x)}{(1+x)(1+x^{2})}dx &=-\frac {35}{64}\zeta (3)+\frac {\pi ^{3}}{32}-\frac {\pi ^{2}\ln 2}{32} \\ \int _{0}^{1}\frac {(1-x)\ln (x)\ln (1-x)}{(1+x)(1+x^{2})}dx &=\frac {7}{64}\zeta (3)-\frac {\pi ^{2}\ln 2}{32} \end {aligned}$$

$$\begin {aligned} \int _{0}^{\infty }\frac {\ln ^{2}x}{(1+x)^2}dx &=\frac {\pi ^{2}}3 \\ \int _{0}^{\infty }\frac {\ln ^{2}x}{(1+x^{2})^{2}}dx &=\frac {\pi ^{3}}{16} \end {aligned}$$

$$\lim _{r\to 0+}\left (\frac {\ln r}2+\int _{0}^{1-r}\frac {dx}{(1-x)(1+x^{2})}\right ) =\frac {\pi }8+\frac {\ln 2}4$$

$$\begin {aligned} \int _{0}^{1}\frac {\ln ^{2}x}{(1-x)(1+x^{2})}dx &=\frac {35}{32}\zeta (3)+\frac {\pi ^{3}}{32} \end {aligned}$$

$$\int _{0}^{\frac {\pi }2}\ln \left (2\sin \frac {x}2\right )\ln \left (2\cos \frac {x}2\right )dx =-\frac {\pi ^{3}}{48}$$

$$\int _{0}^{\frac {\pi }4}\ln (\sin x)\ln (\cos x)dx =-\frac {\pi ^{3}}{96}+\frac {\pi \ln ^{2}2}4$$

$$\begin {aligned} \int _{0}^{1}\frac {\arcsin \sqrt x}{1+x}dx &=\frac {3\pi \ln 2}2-\pi \ln (1+\sqrt 2) \end {aligned}$$

$$\begin {aligned} \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\frac {dxdydz}{\sqrt {1-x^{2}y^{2}z^{2}}} &=\frac {\pi ^{3}}{48}+\frac {\pi \ln ^{2}2}{4} \end {aligned}$$

$$\int _{0}^{1}\frac {\ln (1+x^{2})}{1+x^{2}}dx =\frac {\pi \ln 2}2-G$$

$$\begin {aligned} \int _{0}^{\frac {\pi }4}\ln ^2\left (\sin x\right )dx &=\frac {\pi ^{3}}{192}+\frac {3\pi \ln ^{2}2}{16}+\frac {\ln (2)\beta (2)}2+\Im \operatorname{Li} _{3}(1+i)\\ \int _{0}^{\frac {\pi }4}\ln ^{2}(\cos x)dx &=\frac {7\pi ^{3}}{192}+\frac {5\pi \ln ^{2}2}{16}-\frac {\ln (2)\beta (2)}2-\Im \operatorname{Li} _{3}(1+i) \end {aligned}$$

$$\int _{0}^{2}\frac {dx}{\sqrt {1+x^{3}}} =\frac {\Gamma \left (\frac {1}3\right )^3}{2^{\frac {4}3}\sqrt 3\pi }$$

$$\int _{0}^{\infty }\arctan (\cosh x)\ln (\tanh x)dx =\pi \left (\operatorname{Li} _{2}\left (1-\sqrt 2\right )-\frac {1}4\operatorname {Li} _2 \left (-3+2\sqrt 2\right ) \right )$$

$$\int _{0}^{\frac {\pi }4}\frac {x\ln \tan x}{\sin 2x}dx =-\frac {\pi ^{3}}{64}$$

$$\int _{0}^{\infty }\ln \left (\frac {1+x^{2}+x^{4}+x^{6}}{1+x^{6}}\right )dx =\pi \left (\sqrt 2-1\right )$$

$$\begin {aligned} \int _{0}^{\infty }\frac {\lfloor x\rfloor}{1+e^{2\pi x}}dx &=\frac {1}{24}-\frac {3\ln 2}{16\pi } \end {aligned}$$

$$\begin {aligned} \int _{0}^{\frac {\pi }2}\arctan \left (2\tan ^{2}x\right )dx &=\pi \arctan \frac {1}2 \end {aligned}$$

$$\begin {aligned} \int _{0}^{\infty }\arctan \frac {\sinh x}{1+\cosh 2x}dx &= \frac {2}3\beta (2) \end {aligned}$$