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

Step 1: Understanding the Concept:
We look for the form $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$.
Step 2: Formula Derivation:
Expand the term: $$\left( \frac{1-x}{1+x^2} \right)^2 = \frac{1 + x^2 - 2x}{(1+x^2)^2} = \frac{1+x^2}{(1+x^2)^2} - \frac{2x}{(1+x^2)^2} = \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2}$$
Step 3: Explanation:
Let $f(x) = \frac{1}{1+x^2}$. Then $f'(x) = \frac{d}{dx}(1+x^2)^{-1} = -(1+x^2)^{-2} \cdot (2x) = -\frac{2x}{(1+x^2)^2}$. This exactly matches our expansion! The integral is $\int e^x [f(x) + f'(x)] \, dx$.
Step 4: Final Answer:
The result is $e^x f(x) + C = \frac{e^{x}}{1 + x^2} + C$. Option (a) is correct.