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$$ \int_0^1 \frac{x^\alpha-1}{\ln x} \ dx = \ln (1+\alpha) $$
\begin{align} \int_0^1 \frac{x^\alpha -1}{\ln x} dx &= -\int_0^{\infty}\frac{e^{-\alpha t}-1}{te^t} dt &(x\mapsto e^{-t})\\ &= -\int_0^\infty \frac{1}{te^t}\int_0^{-\alpha} \frac{\partial}{\partial s}(e^{st})dsdt \\ &=-\int_0^{-\alpha} \int_0^\infty e^{-t}e^{st}dtds \\ &=-\int_0^{-\alpha} {\cal{L}}[e^{st}](1) ds \\ &=-\int_0^{-\alpha} \frac{ds}{1-s} \\ &=\ln(1+\alpha) \end{align}
$$ \int_0^\frac{\pi}{2} \tan^\alpha \theta \ d\theta = \frac{\pi}2 \sec{\frac{\alpha\pi}{2}} $$
\begin{align} \int_0^\frac{\pi}{2}\tan^\alpha \theta \ d \theta &= \int_0^\frac{\pi}{2}\sin^\alpha \theta \cos^{-\alpha} \theta \ d \theta \\ &= \int_0^\frac{\pi}{2}\sin^{2\frac{\alpha+1}{2}-1} \theta \cos^{2\frac{-\alpha+1}{2}-1} \theta \ d \theta \\ &= \frac12 {\rm{B}}\left(\frac{\alpha+1}{2}, \frac{-\alpha+1}{2}\right) \\ &= \frac12 \frac{1}{\Gamma(1)}\Gamma\left(\frac{\alpha+1}{2}\right)\Gamma\left(\frac{-\alpha+1}{2}\right) \\ &= \frac{\pi}{2} \csc \left(\pi\cdot\frac{\alpha+1}{2}\right) \\ &= \frac{\pi}{2} \sec \frac{\alpha\pi}{2} \end{align}
$$ \sum_{n=1}^{\infty} \frac{1}{(n+\alpha)(n+\beta)}=\frac{\psi(\beta+1)-\psi(\alpha+1)}{\beta-\alpha} $$
\begin{align} \sum_{n=1}^{\infty} \frac{1}{(n+\alpha)(n+\beta)}&=\frac{1}{\beta-\alpha}\sum_{n=1}^{\infty} \left(\frac{1}{n+\alpha}-\frac{1}{n+\beta}\right) \\ &=\frac{1}{\beta-\alpha}\left(\sum_{n=0}^{\infty}\frac{1}{n+\alpha+1}-\sum_{m=0}^{\infty}\frac{1}{n+\beta+1}\right) \\ &=\frac{\psi(\beta+1)-\psi(\alpha+1)}{\beta-\alpha} \end{align}
$$ \sum_{n=1}^{\infty} \frac{1}{(n+\alpha)^2}=\psi^{(1)}(\alpha+1) $$