Question

$ \int \frac{e^{-x}}{16 + 9e^{-2x}} \, dx$ is equal to :

• A. $ \frac{16}{9} \tan^{-1} (e^{-x}) + C$
• B. $ - \frac{1}{12} \tan^{-1} \left( \frac{3e^{-x}}{4} \right) + C$
• C. $ \tan^{-1} \left( \frac{e^{-x}}{4} \right) + C$
• D. $ - \frac{1}{3} \tan^{-1} \left( \frac{e^{-x}}{4} \right) + C$

Hint:

For integrals involving rational functions with $e^{-x}$, use substitution to simplify and recognize standard forms of integrals like the inverse tangent.

Answer 1

The Correct Option is B

The integral involves a rational function in terms of $e^{-x}$. To simplify, use the substitution $u = e^{-x}$, so $du = -e^{-x}dx$. This leads to the integral becoming: \[ \int \frac{-du}{16 + 9u^2} \] This matches the standard form for the inverse tangent. Thus, the result is: \[ - \frac{1}{12} \tan^{-1} \left( \frac{3u}{4} \right) + C = - \frac{1}{12} \tan^{-1} \left( \frac{3e^{-x}}{4} \right) + C \] Therefore, the correct answer is $(B)$.