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

$\tan^{-1}(e^x)+c$: $c$ is an arbitrary constant

The correct answer is Option (1) → $\tan^{-1}(e^x)+c$: $c$ is an arbitrary constant

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

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

Put $e^x = \tan \theta \Rightarrow dx = \frac{1}{\tan \theta} \cdot \sec^2 \theta \, d\theta$

So, $e^x \, dx = \sec^2 \theta \, d\theta$

Therefore, $\int \frac{e^x \, dx}{e^{2x} + 1} = \int \frac{\sec^2 \theta \, d\theta}{\tan^2 \theta + 1} = \int d\theta = \theta + C$

Now, since $e^x = \tan \theta \Rightarrow \theta = \tan^{-1}(e^x)$