# 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

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

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1. jose flaminio tellez abril says:

aprender a integrar

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learned to integrate

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