Question

Evaluate the integral \( \displaystyle \int \frac{1}{x\log x}\,dx \).

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

Hint:

A very common integral result: \[ \int \frac{f'(x)}{f(x)}dx = \log|f(x)| + C \] Recognizing this pattern makes many logarithmic integrals easy to solve.

Answer 1

The Correct Option is B

Topic: Indefinite Integration (Calculus)
Step 1: Understanding the Question:
The objective is to find the antiderivative of the function \(\frac{1}{x \log x}\).
The structure of the integrand suggests that one part of the function is the derivative of another part.
Step 2: Key Formula or Approach:
We use the Method of Substitution.
If we identify a function \(g(x)\) and its derivative \(g'(x)\) in the integral, we let \(u = g(x)\).
Standard formula: \[ \int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C \]
Step 3: Detailed Explanation:
Let \(u = \log x\).
Differentiating both sides with respect to \(x\):
\[ \frac{du}{dx} = \frac{1}{x} \implies du = \frac{1}{x} dx \]
Now, substitute \(u\) and \(du\) into the original integral:
\[ \int \frac{1}{\log x} \cdot \frac{1}{x} dx = \int \frac{1}{u} du \]
Integrating with respect to \(u\):
\[ \int \frac{1}{u} du = \log|u| + C \]
Substituting back the value of \(u = \log x\):
\[ \log|\log x| + C \]
Step 4: Final Answer:
The integral evaluates to \(\log|\log x| + C\).