Question

Let \( f(x) = \int \frac{e^{3x}}{4 + 8e^{2x} + e^{4x}} \, dx \), and \( g(x) = \int \frac{2\, dx}{e^{3x} + 8e^x + 4e^{-x}} \), then \( f(x) - g(x) = \) ?

• A. \( \frac{1}{2} \tan^{-1} \left( \frac{e^x + 2e^{-x}}{2} \right) + C \)
• B. \( \frac{1}{2} \tan^{-1} \left( \frac{e^x + e^{-x}}{2} \right) + C \)
• C. \( \frac{1}{2} \tan^{-1} \left( \frac{2e^{-x} + e^{2x}}{2} \right) + C \)
• D. \( \frac{1}{2} \tan^{-1} \left( \frac{e^{2x} + 2e^x}{2e^{-x}} \right) + C \)

Hint:

Use substitution \( e^x = t \) to convert exponentials into rational expressions and integrate.

Answer 1

The Correct Option is A

Let’s simplify both integrals: Let \( I = f(x) - g(x) = \int \left( \frac{e^{3x}}{4 + 8e^{2x} + e^{4x}} - \frac{2}{e^{3x} + 8e^x + 4e^{-x}} \right) dx \)
Use substitution: Let \( e^x = t \), then: \[ \begin{align} f(x) = \int \frac{t^3}{4 + 8t^2 + t^4} \cdot \frac{dt}{t} = \int \frac{t^2}{(t^2 + 2)^2} dt \] Similarly transform \( g(x) \). Eventually both simplify and their difference becomes: \[ \begin{align} \frac{1}{2} \tan^{-1} \left( \frac{e^x + 2e^{-x}}{2} \right) + C \]