Question

Evaluate \[ \int \frac{\log x - 1}{1 + (\log x)^2} \, dx \]

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

Hint:

For integrals involving logarithmic functions, use substitution and recognize standard integral forms.

Answer 1

The Correct Option is B

Step 1: Use substitution.
Let \( u = \log x \), so that \( du = \frac{1}{x} dx \). The integral becomes: \[ \int \frac{u - 1}{1 + u^2} \, du. \]
Step 2: Break the integral.
We can separate the terms in the numerator: \[ \int \frac{u}{1 + u^2} \, du - \int \frac{1}{1 + u^2} \, du. \]
Step 3: Perform the integration.
The first integral is \( \frac{1}{2} \log(1 + u^2) \), and the second integral is \( \tan^{-1} u \). Substituting \( u = \log x \), we get the final answer: \[ \frac{x}{1 + (\log x)^2} + c. \]
Step 4: Conclusion.
Thus, the correct answer is \( \frac{x}{1 + (\log x)^2} + c \), corresponding to option (B).