ln-e-x-1-e-x-1-dx-

Question Number 87769 by john santu last updated on 06/Apr/20

\int \frac{\ln (e^{x} + 1)}{e^{- x} + 1} dx

Answered by TANMAY PANACEA. last updated on 06/Apr/20

\int \frac{e^{x} l n (e^{x} + 1)}{e^{x} + 1} d x

t = l n (e^{x} + 1) \to \frac{d t}{d x} = \frac{e^{x}}{e^{x} + 1}

\int t d t = \frac{t^{2}}{2} + c

\frac{{l n (e^{x} + 1)}^{2}}{2} + c

Answered by john santu last updated on 06/Apr/20