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$\int \frac{d x}{e^x+e^{-x}}$ is equal to
(a) $\tan ^{-1}\left(e^x\right)+c$
(b) $\tan ^{-1}\left(e^{-x}\right)+c$
(c) $\log \left(\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}\right)+\mathrm{c}$
(d) $\log \left(\mathrm{e}^x+\mathrm{e}^{-x}\right)+c$
$\int \frac{d x}{e^x+e^{-x}}$ is equal to
(a) $\tan ^{-1}\left(e^x\right)+c$
(b) $\tan ^{-1}\left(e^{-x}\right)+c$
(c) $\log \left(\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}\right)+\mathrm{c}$
(d) $\log \left(\mathrm{e}^x+\mathrm{e}^{-x}\right)+c$
Solution:
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Verified Answer
(a) Let $\mathrm{I}=\int \frac{d x}{e^x+e^{-x}}=\int \frac{e^x d x}{e^{2 x}+1}$;
Put $\mathrm{e}^{\mathrm{x}}=\mathrm{t}, \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$
$\therefore \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^2+1}=\tan ^{-1} \mathrm{t}+\mathrm{c}=\tan ^{-1}\left(\mathrm{e}^{\mathrm{x}}\right)+\mathrm{c} .$
Put $\mathrm{e}^{\mathrm{x}}=\mathrm{t}, \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$
$\therefore \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^2+1}=\tan ^{-1} \mathrm{t}+\mathrm{c}=\tan ^{-1}\left(\mathrm{e}^{\mathrm{x}}\right)+\mathrm{c} .$
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