Integrate $\int \frac{e^x}{9+e^x}dx$ using u-substituion

Question

5.3

Integrate $\int \frac{e^x}{9+e^x}dx$ using u-substituion

Can somebody verify this for me? Thanks!

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Let $u=9+e^x$. Then $\frac{du}{dx}=e^x$ and so $\frac{du}{e^x}=dx$

Thus we have:

$\int \frac{e^x}{9+e^x}dx$

$=\int \frac{e^x}{u}\frac{du}{e^x}$

$=\int \frac{1}{u} du$

$= \ln|u| +C$

$= \ln|9+e^x| +C$

$= \ln(9+e^x) +C$

Where in the last equality we dropped the absolute value because we know $9+e^x$ is always positive.

Answer 1

You are correct, but I feel it is good practice to fully substitute $x$ with $u$. I'll demonstrate:

We let $u=9+e^x\to \frac{du}{dx}=e^x=u-9\implies dx=\frac{du}{u-9}$. Then:

$$\int\frac{e^x}{9+e^x}dx=\int\frac{u-9}{u}\frac{du}{u-9}=\int\frac{du}{u}$$ and then you got the rest.

Answer 2

It's correct. Another custom made integral, in which the denominator of the integrand has its derivative in the numerator. Therefore, by the chain rule, it's $\ln(9+e^x)+C$.