How do you integrate $\int \frac{e^{x}}{4 - e^{x}}$ $int e^x/(4-e^x)$ using substitution?

Question

7427 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-25T11:08:49

$- ln | 4 - e^{x} | + C$ $-lnabs(4-e^x)+C$

$\int \frac{e^{x}}{4 - e^{x}} d x$ $inte^x/(4-e^x)dx$

Let $u = 4 - e^{x}$ $u=4-e^x$ . Differentiating this shows that $d u = - e^{x} d x$ $du=-e^xdx$ . Thus:

$\int \frac{e^{x}}{4 - e^{x}} d x = - \int \frac{- e^{x} d x}{4 - e^{x}} = - \int \frac{d u}{u}$ $inte^x/(4-e^x)dx=-int(-e^xdx)/(4-e^x)=-int(du)/u$

This is a common integral:

$= \int \frac{d u}{u} = - ln | u | + C = - ln | 4 - e^{x} | + C$ $=int(du)/u=-lnabsu+C=-lnabs(4-e^x)+C$

Answer 2 · 2016-10-25T11:48:42

$= - ln (4 - e^{x}) + C$ $=-ln(4-e^x)+C$

Use the substitution $u = 4 - e^{x}$ $u=4-e^x$
Then $d u = - e^{x} d x$ $du=-e^xdx$

So $\int \frac{e^{x} d x}{4 - e^{x}} = \int \frac{- d u}{u}$ $int(e^xdx)/(4-e^x)=int(-du)/u$

$= - ln u$ $=-lnu$

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

How do you integrate ∫ex4−exint e^x/(4-e^x) using substitution?