Question

\(\int \log x^2 \, dx \)

• A. \( \frac{1}{x^2} + k \)
• B. \( \frac{2}{x} + k \)
• C. \( x \log x - x + k \)
• D. \( 2(x \log x - x) + k \)

Hint:

For integrals involving logarithmic functions, use integration by parts and simplify the logarithmic terms whenever possible.

Answer 1

The Correct Option is D

We are given: \[ I = \int \log x^2 \, dx \] To simplify, recall that \( \log x^2 = 2 \log x \). Hence, the integral becomes: \[ I = \int 2 \log x \, dx \] Now, apply integration by parts. Let: \[ u = \log x, \quad dv = 2 dx \] Then, \[ du = \frac{1}{x} dx, \quad v = 2x \] Using the integration by parts formula \( \int u dv = uv - \int v du \), we get: \[ I = 2x \log x - \int 2x \cdot \frac{1}{x} dx = 2x \log x - \int 2 dx = 2x \log x - 2x + k \] Thus, the correct answer is \( 2(x \log x - x) + k \).