Where $c(u, v)=u^{-\alpha}(1-\alpha) $ is the Marshall-Olkin copula density, we have the following integral: \begin{align} I_{1} &=\iint_{E} c(u, v) \log c(u, v) d u d v \\ &=\int_{0}^{1} \int_{0}^{u^{\frac{\alpha}{\beta}}} u^{-\alpha}(1-\alpha) \log \left(u^{-\alpha}(1-\alpha)\right) d v d u \end{align}
They calculate this integral as follows: $$ \begin{aligned} I_{1} &=\int_{0}^{1} u^{\frac{\alpha}{\beta}-\alpha}(1-\alpha) \log \left((1-\alpha) u^{-\alpha}\right) d u \\ &=(1-\alpha) \log (1-\alpha) \int_{0}^{1} u^{\frac{\alpha}{\beta}-\alpha} d u-\alpha(1-\alpha) \int_{0}^{1} u^{\frac{\alpha}{\beta}-\alpha} \log u d u \\ &=\frac{(1-\alpha) \beta \log (1-\alpha)}{\alpha-\alpha \beta+\alpha}+\frac{\alpha(1-\alpha) \beta^{2}}{(\alpha-\alpha \beta+\beta)^{2}} \end{aligned} $$ In the last 3 lines, what are all the steps in-between? I don't understand how the first becomes the second, and the second becomes the last line.
The first equality follows from standard properties of the logarithm, explicitly: $$\log{((1-\alpha)u^{-\alpha})}=\log{(1-\alpha)}+\log{(u^{-\alpha})} = \log{(1-\alpha)}-\alpha\log{(u)}$$ The second follow by the standard integral of $x^{a}$ and integration by parts, explicitly: $$\int_{0}^{1} u^{\frac{\alpha}{\beta}-\alpha} du=\left[\frac{u^{\frac{\alpha}{\beta}-\alpha+1}}{\frac{\alpha}{\beta}-\alpha+1}\right]_{0}^{1}=\frac{\beta}{\alpha-\alpha \beta+\beta}$$ (I think the $+\alpha$ should be a $+\beta$ in the denominator) \begin{align} \int_{0}^{1} u^{\frac{\alpha}{\beta}-\alpha} \log u du &=\left[\frac{u^{\frac{\alpha}{\beta}-\alpha+1}\log u}{\frac{\alpha}{\beta}-\alpha+1}\right]_0^1-\int_{0}^{1}\frac{u^{\frac{\alpha}{\beta}-\alpha}}{\frac{\alpha}{\beta}-\alpha+1}du\\ &=-\left[\frac{u^{\frac{\alpha}{\beta}-\alpha+1}}{(\frac{\alpha}{\beta}-\alpha+1)^2}\right]^1_0\\ &=-\frac{\beta^2}{(\alpha-\alpha \beta+\beta)^2} \end{align} Hope this helps!