I would like to ask this question here too.
The probability density function is \begin{equation} f\left(x;\alpha,\beta,\mu\right)=\alpha\beta\left(1-e^{-\left(x-\mu\right)\beta}\right)^{\alpha-1}e^{-\left(x-\mu\right)\beta},\ x>\mu,\ \alpha>0,\ \beta>0 \end{equation}
I need to show that
\begin{eqnarray*} E\left(\frac{\left(X-\mu\right)e^{-\left(X-\mu\right)\beta}}{1-e^{-\left(X-\mu\right)\beta}}\right) & = & \int_{\mu}^{\infty}\frac{\left(x-\mu\right)e^{-\left(x-\mu\right)\beta}}{1-e^{-\left(x-\mu\right)\beta}}\cdot f\left(x;\alpha,\beta,\mu\right)\enspace dx\\ & = & {\left[\frac{\alpha}{\beta\left(\alpha-1\right)}\left(\psi\left(\alpha\right)-\psi\left(1\right)\right)-\frac{1}{\beta}\left(\psi\left(\alpha+1\right)-\psi\left(1\right)\right)\right]}\tag{1} \end{eqnarray*}
($\psi$ is digamma function)
When i applied the transformation $t=e^{-\left(x-\mu\right)\beta}$ the integral became $$ E\left(\frac{\left(X-\mu\right)e^{-\left(X-\mu\right)\beta}}{1-e^{-\left(X-\mu\right)\beta}}\right)=-\frac{\alpha}{\beta}\int_{0}^{1}t\left(1-t\right)^{\alpha-2}\log t\ dt. $$ how can i go to the next step?
I verified you substituion and got the same answer so we will pick up where you left off. We have $$ I=-\frac{\alpha}{\beta}\int_0^1(\log t) t (1-t)^{\alpha-2}\,\mathrm dt. $$ Using $\partial_x t^x=t^x\log t$ gives $$ I(x)=-\frac{\alpha}{\beta}\partial_x\int_0^1t^x (1-t)^{\alpha-2}\,\mathrm dt, $$ where $I=I(1)$. Notive that $I(x)$ is very closely related to the integral definition of the beta function leaving us with $$ I(x)=-\frac{\alpha}{\beta}\partial_x\frac{\Gamma(\alpha-1)\Gamma(x+1)}{\Gamma(\alpha +x)}. $$ Since $\partial_z\log\Gamma(z):=\psi(z)=\Gamma^\prime(z)/\Gamma(z)\implies\partial_z\Gamma(z)=\psi(z)\Gamma(z)$ we are able to evaluate the derivative yielding $$ I(x)=-\frac{\alpha\Gamma(\alpha-1)}{\beta}\frac{\Gamma(x)}{\Gamma(\alpha+x)}% \left(1+x\psi(x)-x\psi(a+x)\right). $$ Substituting $x=1$ and using the properties $z\Gamma(z)=\Gamma(z+1)$, $\psi(1)=-\gamma$ (Euler-Mascheroni constant), and $H_z=\psi(z+1)-\gamma$ (generalized harmonic numbers) gives after some simplification $$ I=\frac{H_\alpha-1}{(\alpha-1)\beta}. $$ This is clearly different from $(1)$ given in your post. If we use the abovementioned properties we can simplify $(1)$ to $$ \frac{\psi(\alpha)+\frac{1}{\alpha}+\gamma-1}{(\alpha-1) \beta}. $$ Then using the relationship $\psi(z+1)=\psi(z)+\frac{1}{z}$ and the definition of the generalized harmonic numbers shows that the solution for $I$ agrees with the form $(1)$ given in your post.