Solution:

Here,

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

$\rm = {\int \frac{dx}{\frac{(e^x)^2 + 1}{e^x}}}$

$\rm = {\int \frac{e^x dx}{(e^x)^2 \ + 1}} $

Let,

$\rm {e^x\ = \ y} $

Differentiating both sides with respect to x,

$\rm \frac{de^x}{dx} \ = \ \frac{dy}{dx} $

$\rm e^x \ = \ \frac{dy}{dx} $

$\rm \therefore {dy \ = \ e^xdx }$

Putting the values we get,

$\rm = {\int \frac{dy}{y^2 \ + \ 1}}$

$\rm = {\int \frac{dy}{y^2 \ + \ 1^2}}$

$\rm ={ \frac{1}{1} tan^{-^1} \frac{y}{1}} + C $

$\rm ={ tan^{-^1} e^x} + C $

$\rm\therefore {\int \frac{dx}{e^{x} + e^{-x}} = tan^{-1} e^x + C}$