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Question Number 132641 by aurpeyz last updated on 15/Feb/21

$$\int \frac{1}{e^x+9e^{-x}}dx$$

Answered by MJS_new last updated on 15/Feb/21

$$t={\mathrm{e}}^x\to dx=\frac{dt}{t}$$$$\int \frac{dt}{t^2+9}=\frac{1}{3}\arctan \frac{t}{3}=\frac{1}{3}\arctan \frac{{\mathrm{e}}^x}{3}+C$$

Answered by physicstutes last updated on 15/Feb/21

$$\mathcal{I}=\int \frac{1}{e^x+9e^{-x}}dx=\int \frac{e^{x}dx}{e^{2x}+3^2}dx$$$$\mathrm{let}{e}^x=u\Rightarrow du=e^{x}dx$$$$\Rightarrow \mathcal{I}=\int \frac{du}{u^2+3^2}=\frac{1}{3}{\tan }^{-1}\left(\frac{u}{3}\right)+A$$$$\mathcal{I}=\frac{1}{3}{\tan }^{-1}\left(\frac{e^x}{3}\right)+A,A\in \mathbb{R}$$

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