How do you integrate #int e^x/(4-e^x)# using substitution?

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Answer 1 · 2016-10-25T11:08:49

#-lnabs(4-e^x)+C#

#inte^x/(4-e^x)dx#

Let #u=4-e^x#. Differentiating this shows that #du=-e^xdx#. Thus:

#inte^x/(4-e^x)dx=-int(-e^xdx)/(4-e^x)=-int(du)/u#

This is a common integral:

#=int(du)/u=-lnabsu+C=-lnabs(4-e^x)+C#

Answer 2 · 2016-10-25T11:48:42

#=-ln(4-e^x)+C#

Use the substitution #u=4-e^x#
Then #du=-e^xdx#

So #int(e^xdx)/(4-e^x)=int(-du)/u#

#=-lnu#

#-ln(4-e^x)+C#