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

$=-lnu$

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

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