Question

Evaluate the integral \( \int e^x \left( \frac{1 - x}{1 + x^2} \right)^2 dx = \)

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

Hint:

When dealing with integrals of exponential functions and rational functions, apply standard integration rules for each part and use the chain rule if necessary.

Answer 1

The Correct Option is A

Step 1: Apply the chain rule.
We are asked to evaluate the integral \( \int e^x \left( \frac{1 - x}{1 + x^2} \right)^2 dx \). First, we notice that the integrand is the product of an exponential function \( e^x \) and a rational function. Let's simplify and integrate each part.
Step 2: Simplify the expression.
We can rewrite the integrand as: \[ \int e^x \left( \frac{1 - x}{1 + x^2} \right)^2 dx \] This can be seen as a standard form of an integral involving \( e^x \) and the quotient \( \left( \frac{1 - x}{1 + x^2} \right) \).
Step 3: Solve the integral.
The integral simplifies to: \[ e^x \left( \frac{1}{1 + x^2} \right) + C \]
Step 4: Conclusion.
The correct answer is \( e^x \left( \frac{1}{1 + x^2} \right) + C \).