First integral. $X$ has density function $ f(x;\alpha ,\beta ) = \frac{{\beta ^\alpha }}{{\Gamma (\alpha )}}x^{\alpha - 1} e^{ - \beta x},\;\; x > 0, $ with $\alpha,\beta > 0$ fixed. It follows readily from \Gamma '(\alpha ) = \int_0^\infty {x^{\alpha - 1} e^{ - x} \ln x\,{\rm d}x} and \Gamma '(\alpha ) = \Gamma (\alpha )\psi _0 (\alpha ), where $\psi_0$ is the digamma function, that $ {\rm E}[X\ln X] = \frac{{\beta ^\alpha }}{{\Gamma (\alpha )}}\int_0^\infty {(x\ln x)x^{\alpha - 1} e^{ - \beta x} dx} = \frac{\alpha }{\beta }[\psi_0 (\alpha + 1) - \ln \beta ]. $
EDIT: Elaborating (in response to the OP's request).
First, you can find here the above formulas for \Gamma'(\alpha). Now, using a change of variable, we have \Gamma'(\alpha+1) = \int_0^\infty {\beta ^{\alpha+1} x^{\alpha} e^{ - \beta x} \ln (\beta x)\,{\rm d}x} = \ln \beta \int_0^\infty {\beta ^{\alpha+1} x^{\alpha} e^{ - \beta x} \,{\rm d}x} + \int_0^\infty { \beta ^{\alpha+1} x^{\alpha} e^{ - \beta x} \ln x\,{\rm d}x}. The first integral on the right-hand side is equal to $\Gamma(\alpha+1)\ln \beta$, and the second integral to $\Gamma(\alpha)\beta {\rm E}[X\ln X]$. It thus follows that $ \Gamma (\alpha + 1)\psi _0 (\alpha + 1) = \Gamma (\alpha + 1)\ln \beta + \Gamma (\alpha )\beta {\rm E}[X\ln X]. $ Finally, by $\Gamma(\alpha+1)=\alpha \Gamma(\alpha)$, we get $ {\rm E}[X\ln X] = \frac{\alpha }{\beta }[\psi_0 (\alpha + 1) - \ln \beta ]. $