Question:
Evaluate $\int \frac{1}{e^{x}+e^{-x}} d x$
Solution:
$\int \frac{1}{e^{x}+e^{-x}} d x$
We can write above integral as:
$=\int \frac{1}{e^{x}+\frac{1}{e^{x}}} d x$
$=\int \frac{e^{x}}{e^{2 x}+1} d x-(1)$
Let $e^{x}=t$
Differentiating w.r.t $\mathrm{x}$ we get,
$e^{x} d x=d t$
$\therefore$ integral (1) becomes,
$=\tan ^{-1}(\mathrm{t})+C\left(\because \int \frac{1}{x^{2}+1} d x=\tan ^{-1}(x)\right)$
Putting value of $t$ we get,
$=\tan ^{-1}\left(e^{x}\right)+C$
$\therefore \int \frac{1}{e^{x}+e^{-x}} d x=\tan ^{-1}\left(e^{x}\right)+C$
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