Question Number 85468 by M±th+et£s last updated on 22/Mar/20
$$\int{ln}\left(\mathrm{1}−{e}^{{x}} \right)\:{dx} \\ $$
Answered by mind is power last updated on 22/Mar/20
$$={xln}\left(\mathrm{1}−{e}^{{x}} \right){dx}+\int\frac{{xe}^{{x}} }{\mathrm{1}−{e}^{{x}} }{dx} \\ $$$$={xln}\left(\mathrm{1}−{e}^{{x}} \right)+\int\left(−{x}+\frac{{x}}{\mathrm{1}−{e}^{{x}} }\right){dx} \\ $$$$={xln}\left(\mathrm{1}−{e}^{{x}} \right)−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\int\frac{{x}}{\mathrm{1}−{e}^{{x}} }{dx} \\ $$$$\mathrm{1}−{e}^{{x}} ={y}\Rightarrow \\ $$$$={xln}\left(\mathrm{1}−{e}^{{x}} \right)−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\int\frac{{ln}\left(\mathrm{1}−{y}\right)}{{y}}.\frac{{dy}}{\left(\mathrm{1}−{y}\right)} \\ $$$$\int\frac{{ln}\left(\mathrm{1}−{y}\right)}{\left(\mathrm{1}−{y}\right){y}}{dy}=\int\frac{{ln}\left(\mathrm{1}−{y}\right)}{{y}\left(\mathrm{1}−{y}\right)}{dy}=\int\frac{{ln}\left(\mathrm{1}−{y}\right){dy}}{{y}}+\int\frac{{ln}\left(\mathrm{1}−{y}\right)}{\mathrm{1}−{y}}{dy} \\ $$$$=−{Li}_{\mathrm{2}} \left({y}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}−{y}\right)+{c} \\ $$$$={xln}\left(\mathrm{1}−{e}^{{x}} \right)−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{Li}_{\mathrm{2}} \left(\mathrm{1}−{e}^{{x}} \right)−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$$$={xln}\left(\mathrm{1}−{e}^{{x}} \right)−{Li}_{\mathrm{2}} \left(\mathrm{1}−{e}^{{x}} \right)−{x}^{\mathrm{2}} +{c} \\ $$
Commented by M±th+et£s last updated on 22/Mar/20
$${god}\:{bless}\:{you}\:{sir}\:{thank}\:{you} \\ $$
Commented by mind is power last updated on 22/Mar/20
$${happy}\:{To}\:{help} \\ $$