Question

The integral \( \int \frac{dx}{e^x + e^{-x} + 2} \) is equal to:

• A. \( \frac{1}{e^x + 1} + c \)
• B. \( \frac{-1}{e^x + 1} + c \)
• C. \( \frac{1}{1 + e^{-x}} + c \)
• D. \( \frac{1}{e^{-x} - 1} + c \)
• E. \( \frac{1}{e^x - 1} + c \)

Hint:

Exponential integrals involving both positive and negative powers are almost always solved by "cleaning up" the denominator first. Multiplying by \( e^x \) is a standard move to convert the problem into a simple rational function.

Answer 1

The Correct Option is B

Concept: To integrate expressions with \( e^x \) and \( e^{-x} \), it is usually best to transform the integrand into a form where substitution is possible. Multiplying the numerator and denominator by \( e^x \) helps eliminate negative exponents and often reveals a perfect square or a clear \( u \)-substitution path.

Step 1: Transforming the integrand.
Multiply the numerator and denominator by \( e^x \): \[ \int \frac{e^x \, dx}{e^{2x} + 1 + 2e^x} \] Notice that the denominator is a perfect square trinomial: \( (e^x)^2 + 2(e^x) + 1 = (e^x + 1)^2 \). \[ = \int \frac{e^x}{(e^x + 1)^2} \, dx \]

Step 2: Applying \( u \)-substitution.
Let \( u = e^x + 1 \). Then \( du = e^x \, dx \). The integral becomes: \[ \int \frac{du}{u^2} = \int u^{-2} \, du \] \[ = \frac{u^{-1}}{-1} + c = -\frac{1}{u} + c \] Substituting back for \( u \): \[ = -\frac{1}{e^x + 1} + c \]