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
$$
{\rm E}[X\ln X] = \frac{\alpha }{\beta }[\psi_0 (\alpha + 1) - \ln \beta ].
$$