Question

$\int \frac{dx}{e^ x+e^{-x}} \, dx$ is equal to,
(A) $tan^{-1} (e^x) + C$
(B) $tan^{-1} (e^{-x}) + C$
(C) $log (e^x - e^{-x}) + C$
(D) $log (e^x + e^{-x}) + C$

Answer 1

check the attachment

Answer 2

Answer:

Option B is correct.

Step-by-step explanation:

To integrate the function

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

$\int \frac{1}{e^ x+e^{-x}} \, dx \\ \\ \int \frac{1}{e^ x+ \frac{1}{ {e}^{x} }} \, dx \\ \\ \int \frac{1}{ \frac{ {e}^{2x} + 1}{ {e}^{x} }} \, dx \\ \\ \int \frac{ {e}^{x} }{e^ {2x}+1} \, dx \\ \\ put \: {e}^{x} = t \\ \\ {e}^{x} dx = dt \\ \\ substitute \: these \: values \\ \\ \int \frac{ 1 }{t^ {2}+1} \, dt \\ \\ \because \: \int \: \frac{1}{ {x}^{2} + 1} dx = tan ^{ - 1} \: x + C \\ \\ so \\ \\ \int \frac{ 1 }{t^ {2}+1} \, dt = tan ^{ - 1} \: t + C \\ \\ \int \frac{1}{e^ x+e^{-x}} \, dx = tan ^{ - 1} \: {e}^{x} + C$

Option B is correct.

Hope it helps you.