Problem 2-8: Integration

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


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)
Problem 2-8

3 thoughts on “Problem 2-8: Integration

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