Question Number 93553 by Ar Brandon last updated on 13/May/20
$$\int{ln}\left(\mathrm{1}+\mathrm{e}^{\mathrm{u}} \right)\mathrm{du} \\ $$
Commented by Tony Lin last updated on 13/May/20
$${let}\:−{e}^{{u}} ={z}\:\:{dz}={zdu} \\ $$$$\int\:\:\frac{{ln}\left(\mathrm{1}−{z}\right)}{{z}}{dz} \\ $$$$=−{Li}_{\mathrm{2}} \left({z}\right)+{c} \\ $$$$=−{Li}_{\mathrm{2}} \left(−{e}^{{u}} \right)+{c} \\ $$$${Li}_{\mathrm{2}} \left({z}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{z}^{{k}} }{{k}^{\mathrm{2}} } \\ $$
Commented by Ar Brandon last updated on 13/May/20
Okay thanks ��