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Question Number 76467 by kaivan.ahmadi last updated on 27/Dec/19

$$\int \frac{dx}{\sqrt{e^{2x}-4}\left({sec}^{-1}\left(\frac{e^x}{4}\right)\right)}$$

Answered by john santu last updated on 28/Dec/19

$$let:\sec u=\frac{e^x}{4}\to e^{x}dx=4\sec u\times \tan udu$$$$I=\int \frac{4\sec u\times \tan udu}{e^x\left(\sqrt{e^{2x}-4}\right)}$$$$=\int \frac{4\sec u.\tan udu}{4\sec u\sqrt{16\sec {}^{2}u-4}}$$$$\int \frac{\tan udu}{2\sqrt{4\sec {}^{2}u-1}}=\int \frac{\tan udu}{2\sqrt{4\tan {}^{2}u+3}}$$$$nowlet\tan u=\beta$$

Commented by kaivan.ahmadi last updated on 29/Dec/19

$$thankyousir$$

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