Expectation about Generalized exponential distribution

I would like to ask this question here too.

The probability density function is $$f\left(x;\alpha,\beta,\mu\right)=\alpha\beta\left(1-e^{-\left(x-\mu\right)\beta}\right)^{\alpha-1}e^{-\left(x-\mu\right)\beta},\ x>\mu,\ \alpha>0,\ \beta>0$$

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]} \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?