How do you find the antiderivative of #e^(2x)/sqrt(1-e^x)#?

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Answer 1 · 2016-09-09T21:55:36

#(-2(e^x+2)sqrt(1-e^x))/3+C#

We have:

#I=inte^(2x)/sqrt(1-e^x)dx#

Let #u=e^x#. This implies that #du=e^xdx#. We can write #e^(2x)# as #e^x(e^x)#:

#I=int(e^x(e^x)dx)/sqrt(1-e^x)=intu/sqrt(1-u)du#

Letting #v=1-u#, such that #dv=-du#, and manipulating to show that #u=1-v#:

#I=-int(1-v)/sqrtvdv=int(v^(1/2)-v^(-1/2))dv#

Integrating using the #intv^ndv=v^(n+1)/(n+1),n!=-1# rule:

#I=v^(3/2)/(3/2)-v^(1/2)/(1/2)=2/3v^(3/2)-2v^(1/2)=(2sqrtv(v-3))/3#

Since #v=1-u#, and #u=e^x#, so #v=1-e^x#:

#I=(2sqrt(1-e^x)(1-e^x-3))/3=(-2(e^x+2)sqrt(1-e^x))/3+C#