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

The integral is $\int \frac{dx}{e^x + e^{-x}}$

Multiply both numerator and denominator by $e^x$:

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

$= \int \frac{e^x \cdot dx}{e^x(e^x + e^{-x})}$

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Expand the denominator:

$= \int \frac{e^x \, dx}{e^x \cdot e^x + e^x \cdot e^{-x}}$

$= \int \frac{e^x \, dx}{e^{2x} + 1}$

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Let $t = e^x$

Then $\frac{dt}{dx} = e^x$

So $dt = e^x \, dx$

Substituting:

$\int \frac{e^x \, dx}{e^{2x} + 1}$

$= \int \frac{dt}{t^2 + 1}$

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Using the standard integral formula:

$\int \frac{dt}{t^2 + 1} = \tan^{-1}(t) + c$

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Substituting back $t = e^x$:

$\tan^{-1}(e^x) + c$