Evaluate $\int\limits_{0}^{1} \frac{dx}{e^x + e^{-x}}$

$\tan^{-1} e - \frac{\pi}{4}$

The correct answer is Option (1) → $\tan^{-1} e - \frac{\pi}{4}$

Let $I = \int\limits_{0}^{1} \frac{dx}{e^x + e^{-x}} = \int\limits_{0}^{1} \frac{e^x}{1 + e^{2x}} dx$

[Divide by $e^{-x}$ in both numerator and denominator]

Put $e^x = t$

$\Rightarrow e^x dx = dt$

For limits, when $x = 0$, then $e^0 = t \Rightarrow t = 1$

and when $x = 1$, then $e^1 = t \Rightarrow t = e$

$∴I = \int\limits_{1}^{e} \frac{dt}{1 + t^2} = [\tan^{-1} t]_1^e \left[ ∵\int \frac{1}{1 + t^2} dt = \tan^{-1} t \right]$

$= \tan^{-1} e - \tan^{-1} 1  \left[ ∵\tan \frac{\pi}{4} = 1 \right]$

$= \tan^{-1} e - \frac{\pi}{4}$