Question

\( \int e^x \left(\frac{1-x}{1+x^2}\right)^2 dx =\) _____ + C

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

Hint:

Whenever you see \( e^x \) multiplied by a rational function, try checking if it matches derivative of a quotient.

Answer 1

The Correct Option is A

Concept: Use reverse product rule: \[ \frac{d}{dx}\left(\frac{e^x}{1+x^2}\right) \] Step 1: Differentiate RHS. \[ \frac{d}{dx}\left(\frac{e^x}{1+x^2}\right) = \frac{e^x(1+x^2) - e^x(2x)}{(1+x^2)^2} \] \[ = \frac{e^x(1 + x^2 - 2x)}{(1+x^2)^2} \] Step 2: \[ = \frac{e^x(1 - x)^2}{(1+x^2)^2} \] Step 3: \[ \int e^x \left(\frac{1-x}{1+x^2}\right)^2 dx = \frac{e^x}{1+x^2} + C \]