ln-1-e-x-dx-

Question Number 85468 by M±th+et£s last updated on 22/Mar/20

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

Answered by mind is power last updated on 22/Mar/20

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

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

= x l n (1 - e^{x}) - \frac{x^{2}}{2} + \int \frac{x}{1 - e^{x}} d x

1 - e^{x} = y \Rightarrow

= x l n (1 - e^{x}) - \frac{x^{2}}{2} + \int \frac{l n (1 - y)}{y} . \frac{d y}{(1 - y)}

\int \frac{l n (1 - y)}{(1 - y) y} d y = \int \frac{l n (1 - y)}{y (1 - y)} d y = \int \frac{l n (1 - y) d y}{y} + \int \frac{l n (1 - y)}{1 - y} d y

= - {L i}_{2} (y) - \frac{1}{2} {l n}^{2} (1 - y) + c

= x l n (1 - e^{x}) - \frac{x^{2}}{2} - {L i}_{2} (1 - e^{x}) - \frac{x^{2}}{2} + c

= x l n (1 - e^{x}) - {L i}_{2} (1 - e^{x}) - x^{2} + c

Commented by M±th+et£s last updated on 22/Mar/20

g o d b l e s s y o u s i r t h a n k y o u

Commented by mind is power last updated on 22/Mar/20

h a p p y T o h e l p