Question

The integral \( \int e^x \left(\tan^{-1}x + \frac{1}{1+x^2}\right) dx \) is equal to

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

Hint:

Always look for the pattern \( \int e^x (f(x) + f'(x)) dx \) when you see an \( e^x \) multiplied by a sum of functions. Identifying \( f(x) \) and its derivative \( f'(x) \) simplifies the integral significantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks to evaluate an indefinite integral that has the form \( \int e^x (f(x) + f'(x)) dx \).

Step 2: Key Formula or Approach:
The standard integral formula for this form is:
\[ \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \]

Step 3: Detailed Explanation:
Given integral: \( I = \int e^x \left(\tan^{-1}x + \frac{1}{1+x^2}\right) dx \)
Identify \( f(x) \) and \( f'(x) \):
Let \( f(x) = \tan^{-1}x \).
Then, the derivative of \( f(x) \) is \( f'(x) = \frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2} \).
The integral is exactly in the form \( \int e^x (f(x) + f'(x)) dx \).
Applying the formula:
\[ I = e^x \tan^{-1}x + C \]

Step 4: Final Answer:
The integral is equal to \( e^x \tan^{-1}x + C \).