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

Answer 1

William Abdalla

#=-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#

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Answer 2

Emily Ammerman

#-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#

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Answer 3

Cameron Alexander

To integrate ( \frac{e^x}{4 - e^x} ) using substitution, let ( u = 4 - e^x ). Then, ( du = -e^x dx ).

Now substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^x}{4 - e^x} dx = -\int \frac{1}{u} du ]

This simplifies to:

[ -\ln|u| + C ]

Substitute back for ( u ) to get the final answer:

[ -\ln|4 - e^x| + C ]

where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.