Question

If \( \int \frac{1}{2x^2} \, dx = k \cdot 2x + C \), then \( k \) is equal to:

• A. \( -1 \)
• B. \( \log 2 \)
• C. \( -\log 2 \)
• D. \( 1/2 \)

Hint:

For integrals involving powers of \( x \), use the power rule: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).

Answer 1

The Correct Option is D

The integral of \( \frac{1}{2x^2} \) is \( \int \frac{1}{2x^2} \, dx = -\frac{1}{2x} + C \). Thus, comparing with the given form \( k \cdot 2x + C \), we find that \( k = \frac{1}{2} \).