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Question Number 168952 by Sotoberry last updated on 22/Apr/22

Answered by alephzero last updated on 22/Apr/22

$$\int \frac{2+e^x}{e^x}dx$$$$u=e^x$$$$du=e^{x}dx$$$$dx=\frac{du}{e^x}=\frac{du}{u}$$$$\Rightarrow \int \frac{2+e^x}{e^x}dx=\int \frac{2+u}{u}\frac{du}{u}=$$$$=\int \frac{2+u}{u^2}du=\int \frac{2}{u^2}du+\int \frac{1}{u}du=$$$$=2\int u^{-2}du+\ln \mid u\mid =$$$$=2(-1u^{-1})+\ln \mid u\mid +c$$$$=-\frac{2}{u}+\ln \mid u\mid +c=-\frac{2}{e^x}+\ln \mid e^x\mid +c$$$$\mathrm{\forall }x\in \mathbb{R}{e}^x>0$$$$\Rightarrow -\frac{2}{e^x}+\ln \mid e^x\mid +c=-\frac{2}{e^x}+x+c$$$$\Rightarrow \int \frac{(2+e^x)dx}{e^x}=x-\frac{2}{e^x}+c$$

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