# Problem 2-8: Integration

Find $\displaystyle\int \! \frac{e^x}{1+e^x}\, dx$ .

Solution

Let $u=1+e^x$ , therefore $du=e^x dx \to dx=\frac{1}{u-1}du$ . Hence

$\displaystyle\int \! \frac{e^x}{1+e^x}\, dx = \displaystyle\int \! \frac{u-1}{u} \frac{1}{u-1}\,du = \displaystyle\int \! \frac{1}{u} \,du=\ln (u)=\ln(1+e^x)$

## 3 thoughts on “Problem 2-8: Integration”

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