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$\int \frac{d x}{e^{x}+e^{-x}}$ is equal to

A. $\tan ^{-1}\left(e^{x}\right)+\mathrm{C}$

B. $\tan ^{-1}\left(e^{-x}\right)+\mathrm{C}$

C. $\log \left(e^{x}-e^{-x}\right)+\mathrm{C}$

D. $\log \left(e^{x}+e^{-x}\right)+\mathbf{C}$


Let $I=\int \frac{d x}{e^{x}+e^{-x}} d x=\int \frac{e^{x}}{e^{2 x}+1} d x$

Also, let $e^{x}=t \Rightarrow e^{x} d x=d t$

$\therefore I=\int \frac{d t}{1+t^{2}}$

$=\tan ^{-1} t+\mathrm{C}$

$=\tan ^{-1}\left(e^{x}\right)+\mathrm{C}$

Hence, the correct answer is A.


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