Question

If \( \int \log x \, dx = x \log x + k(x) + c \) then

• A. \( k(x) = \log x \)
• B. \( k(x) = -\log x \)
• C. \( k(x) = -x \)
• D. \( k(x) = -x^2 \)

Hint:

The integral of \( \log x \) is a standard result that is very useful to memorize for competitive exams: \( \int \log x \, dx = x \log x - x + c \). Knowing this can save you the time of performing integration by parts.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question requires evaluating the indefinite integral of \( \log x \) and then comparing the result with the given expression to find the function \( k(x) \).
Step 2: Key Formula or Approach:
We will use the method of integration by parts to solve \( \int \log x \, dx \). The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \] Step 3: Detailed Explanation:
To evaluate \( \int \log x \, dx \), we can write it as \( \int (\log x) \cdot 1 \, dx \). Let's choose our u and dv according to the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential). Let \( u = \log x \) and \( dv = 1 \, dx \).
Then, we find du and v: \[ du = \frac{1}{x} \, dx \] \[ v = \int 1 \, dx = x \] Now, applying the integration by parts formula: \[ \int \log x \, dx = (\log x)(x) - \int x \left(\frac{1}{x}\right) \, dx \] \[ \int \log x \, dx = x \log x - \int 1 \, dx \] \[ \int \log x \, dx = x \log x - x + c \] Now, we compare this result with the given equation: \[ x \log x - x + c = x \log x + k(x) + c \] By direct comparison of the terms, we can see that: \[ k(x) = -x \] Step 4: Final Answer:
The function \( k(x) \) is \( -x \). Therefore, the correct option is (iii).