Question

\( \int \left( \frac{1-\log x}{1+(\log x)^2} \right)^2 dx = \)

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

Hint:

When \( \log x \) appears in a rational function within an integral, substituting \( t = \log x \) usually reveals an \( e^t(f+f') \) structure.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

Substitute \( \log x = t \) to transform the integral into a standard form involving \( e^t \).
Step 2: Key Formula or Approach:

1. Substitution: \( x = e^t \implies dx = e^t dt \). 2. Standard Integral: \( \int e^t [f(t) + f'(t)] dt = e^t f(t) + c \).
Step 3: Detailed Explanation:

Let \( t = \log x \). Then \( I = \int \left( \frac{1-t}{1+t^2} \right)^2 e^t dt \). Expand the term inside the integral: \[ \left( \frac{1-t}{1+t^2} \right)^2 = \frac{1+t^2-2t}{(1+t^2)^2} = \frac{1+t^2}{(1+t^2)^2} - \frac{2t}{(1+t^2)^2} \] \[ = \frac{1}{1+t^2} + \left( -\frac{2t}{(1+t^2)^2} \right) \] Let \( f(t) = \frac{1}{1+t^2} \). Then \( f'(t) = \frac{d}{dt}(1+t^2)^{-1} = -(1+t^2)^{-2}(2t) = -\frac{2t}{(1+t^2)^2} \). The integral becomes: \[ \int e^t [f(t) + f'(t)] dt = e^t f(t) + c \] \[ = e^t \cdot \frac{1}{1+t^2} + c \] Substitute back \( t = \log x \) and \( e^t = x \): \[ = \frac{x}{1+(\log x)^2} + c \]
Step 4: Final Answer:

Matches Option (C).