Question

$$ \int e^x \left( \log x + \frac{1}{x^2} \right) dx = ? $$

• A. $e^x \left\{ \log x - \frac{1}{x} \right\} + c$
• B. $e^x \left( \log x + \frac{1}{x} \right) + c$
• C. $e^x \{ \log x \} + c$
• D. None

Hint:

When an integrand with $e^x$ doesn't immediately show a function and its derivative, look for "hidden" pairs by adding and subtracting terms like $1/x$, $1/x^2$, or trigonometric functions.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem utilizes the special integral form $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$. We need to manipulate the terms inside the parentheses to fit this structure.

Step 2: Key Formula or Approach:
Identify a function $f(x)$ such that the integrand becomes $e^x (f(x) + f'(x))$. Since $\frac{1}{x^2}$ is not the direct derivative of $\log x$, we add and subtract $\frac{1}{x}$ inside the bracket.

Step 3: Detailed Explanation:
1. Rewrite the integral by adding and subtracting $\frac{1}{x}$: \[ \int e^x \left( \log x + \frac{1}{x} - \frac{1}{x} + \frac{1}{x^2} \right) dx \] 2. Group the terms: \[ \int e^x \left[ \left( \log x - \frac{1}{x} \right) + \left( \frac{1}{x} + \frac{1}{x^2} \right) \right] dx \] 3. Let $f(x) = \log x - \frac{1}{x}$. 4. Find the derivative $f'(x)$: \[ f'(x) = \frac{d}{dx}(\log x) - \frac{d}{dx}\left(\frac{1}{x}\right) = \frac{1}{x} - \left(-\frac{1}{x^2}\right) = \frac{1}{x} + \frac{1}{x^2} \] 5. The integral is now in the form $\int e^x [f(x) + f'(x)] \, dx$. 6. Applying the rule: \[ e^x f(x) + c = e^x \left( \log x - \frac{1}{x} \right) + c \]

Step 4: Final Answer
The integral is $e^x \left( \log x - \frac{1}{x} \right) + c$.