Derivation of a log of an integral I've come across the following derivation
$$
\frac{\partial}{\partial \alpha}  \log_2 \left(\int p(x)^\alpha \, \mathrm{d}x \right)
$$
but cannot figure out, what tricks are required to solve it nicely. Definitely some chain rule but the log of the integral looks scary.
($p(x)$ is probability density; an alternative version with probability mass function, with sum instead of integral, was easy)
 A: $$\frac{\partial \log _2\left(\int p(x)^{\alpha } \, dx\right)}{\partial \alpha }=\frac{\partial \frac{\log \left(\int p(x)^{\alpha } \, dx\right)}{\log 2}}{\partial \alpha }=\frac{1}{\log 2}\frac{\partial \log \left(\int p(x)^{\alpha } \, dx\right)}{\partial \alpha }=$$
$$=\frac{1}{\log 2}\frac{\int p(x)^{\alpha } \log (p(x)) \, dx}{\int p(x)^{\alpha } \, dx}$$

Example
$$\log _2\left(\int x^{\alpha } \, dx\right)=\frac{\log \left(\frac{x^{\alpha +1}}{\alpha +1}\right)}{\log 2}+C$$
$$\frac{\partial}{\partial\alpha}\log _2\left(\int x^{\alpha } \, dx\right)=\frac{\partial}{\partial\alpha}\frac{\log \left(\frac{x^{\alpha +1}}{\alpha +1}\right)}{\log 2}=\frac{1}{\log 2}\frac{\partial}{\partial\alpha}\left((\alpha+1)\log x-\log(\alpha+1)\right)=\frac{1}{\log 2}\left(\log x-\frac{1}{\alpha+1}\right)$$
Using the formula above
$$\frac{\partial}{\partial\alpha}\log _2\left(\int x^{\alpha } \, dx\right)=\frac{1}{\log 2}\frac{\int \left(x^{\alpha}\log x\right)\,dx}{\int x^{\alpha}\,dx}=$$
$$=\frac{1}{\log 2}\frac{\frac{x^{\alpha +1} ((\alpha +1) \log x-1)}{(\alpha +1)^2}}{\frac{x^{\alpha +1}}{\alpha +1}}=\frac{1}{\log 2}\frac{x^{\alpha +1} ((\alpha +1) \log x-1)}{(\alpha +1)^2}\cdot \frac{\alpha +1}{x^{\alpha +1}}=\frac{1}{\log 2}\frac{(\alpha +1) \log (x)-1}{\alpha +1}$$
