# About Laplace transform

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}

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Where did the $s$ come from? – Ron Gordon Feb 4 '13 at 3:46
@rlgordonma I dont get it as well :( – Ryan Feb 4 '13 at 4:06
@Ryan: is there some bit of information about the problem we are overlooking? Composition of product of functions or something? Regards – Amzoti Feb 4 '13 at 4:51
@Amzoti thats a lemma in solving this integral – Ryan Feb 4 '13 at 5:25
@Ryan: can you write out everything about the problem as there is an important tidbit missing? Regards – Amzoti Feb 4 '13 at 5:29