The integral $\int\frac{e^{-x}}{9+4e^{-2

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Application of Integrals

Question:

The integral $\int\frac{e^{-x}}{9+4e^{-2x}}dx$ is equal to

Options:

$\frac{1}{6}\tan^{-1}\left(\frac{2e^{-x}}{3}\right)+C$

$-\frac{1}{6}\tan^{-1}\left(\frac{2e^{-x}}{3}\right)+C$

$\frac{1}{12}\tan^{-1}\left(\frac{2e^{-x}}{3}\right)+C$

$-\frac{1}{12}\tan^{-1}\left(\frac{2e^{-x}}{3}\right)+C$

Correct Answer:

$-\frac{1}{6}\tan^{-1}\left(\frac{2e^{-x}}{3}\right)+C$

Explanation:

$I=\int\frac{e^{-x}}{9+4e^{-2x}}dx$

let $y=e^{-x}$

$dy=-e^{-x}dx⇒-dy=e^{-x}dx$

$I=-\int\frac{dy}{9+4y^2}⇒I=\frac{-1}{4}\int\frac{dy}{(3/2)^2+y^2}$

$⇒I=\frac{-1}{4}×\frac{2}{3}\tan^{-1}\frac{2y}{3}+C$

$⇒I=-\frac{1}{6}\tan^{-1}\left(\frac{2e^{-x}}{3}\right)+C$