Question

\(∫\frac{dx}{e^x+e^{-x}}\) is equal to

• A. \(tan^{-1}(e^x)+C\)
• B. \(tan^{-1}(e^{-x})+C\)
• C. \(tan^{-1}(e^x-e^{-x})+C\)
• D. \(log(e^x+e^{-x})+C\)

Hint:

Let's go through the steps to solve the integral using the substitution method 

Given:
\[I = \int \frac{dx}{e^x + e^{-x}}\]

First, rewrite the integrand:
\[I = \int \frac{dx}{e^x + e^{-x}} = \int \frac{dx}{e^x \left(1 + e^{-2x}\right)}\]

Now, make the substitution \( e^x = t \). Therefore, \( dx = \frac{dt}{t} \):

Substitute \( e^x = t \) into the integral:
\[I = \int \frac{dt/t}{t(1 + t^{-2})} = \int \frac{dt}{t^2 + 1}\]

This simplifies to:
\[I = \int \frac{dt}{1 + t^2}\]

The integral of \(\frac{1}{1 + t^2}\) is the arctangent function:
\[I = \arctan(t) + C\]

Now, substitute back \( t = e^x \):
\[I = \arctan(e^x) + C\]

Therefore, the integral \(\int \frac{dx}{e^x + e^{-x}}\) is:
\[I = \arctan(e^x) + C\]
Hence, the correct Answer is Option A\(=tan^{-1}(e^x)+C\)

Answer 1

The Correct Option is A

The correct answer is A:\(=tan^{-1}(e^x)+C\)
Let \(I=∫\frac{dx}{e^x+e^{-x}}dx=∫\frac{e^x}{e^{2x}+1}dx\)
Also,let \(e^x=t⇒e^x dx=dt\)
\(∴I=∫\frac{dt}{1+t^2}\)
\(=tan^{-1}+C\)
\(=tan^{-1}(e^x)+C\)
Hence,the correct Answer is A.