About Laplace transform

Question

I dont understand the following working, why the integral becomes double integral? $$\begin{align} & \ \ \ \int_{0}^{1}{{{\left( \frac{1}{\ln x}+\frac{1}{1-x} \right)}^{2}}\text{d}x}=\mathcal{L}\left[ \frac{\left(t-1+{{\text{e}}^{-t}}\right)^2}{{{t}^{2}}} \right]=\int_{s}^{\infty }{\int_{u}^{\infty }{\mathcal{L}\left[ {{\left( v-1+{{\text{e}}^{-v}} \right)}^{2}} \right]}}\text{d}v\text{d}u \\ & =\int_{s}^{\infty }{\int_{u}^{\infty }{\left( \frac{2}{{{v}^{3}}}-\frac{2}{{{v}^{2}}}+\frac{1}{v}+\frac{2}{{{\left( v+1 \right)}^{2}}}-\frac{2}{v+1}+\frac{1}{v+2} \right)\text{d}v\text{d}}}u \\ & =\int_{s}^{\infty }{\left( \frac{1}{{{u}^{2}}}-\frac{1}{u}-\ln u-\frac{2}{u+1}-2\ln \left( u+1 \right)+\ln \left( u+2 \right) \right)}\text{d}u \\ & =\frac{1}{s}+\left( 2+s \right)\left( \ln s+\ln \left( s+2 \right)-2\ln \left( s+1 \right) \right) \\ \end{align}$$