Question

If \( \int \frac{3ax}{b^2 + c^2x^2} dx = A \log |b^2 + c^2x^2| + K \), then the value of \( A \) is

• A. \( 3a \)
• B. \( \frac{3a}{2b^2} \)
• C. \( \frac{3a}{b^2c^2} \)
• D. \( \frac{3a}{2c^2} \)

Hint:

Use the shortcut: \( \int \frac{f'(x)}{f(x)} dx = \log |f(x)| + C \).
Adjust the constant factor in the numerator to match the derivative of the denominator exactly.

Answer 1

The Correct Option is D

Concept:
• Integration by substitution: If an integral has the form \( \int f(g(x))g'(x) dx \), we substitute \( u = g(x) \).
• The derivative of the denominator often appears in the numerator of log-type integrals.

Step 1: Choose an appropriate substitution for the integral
Let the integral be \( I \): \[ I = \int \frac{3ax}{b^2 + c^2x^2} dx \] Put \( t = b^2 + c^2x^2 \). Differentiating with respect to \( x \): \[ \frac{dt}{dx} = 0 + c^2(2x) = 2c^2x \] \[ dt = 2c^2x dx \implies x dx = \frac{1}{2c^2} dt \]

Step 2: Perform the substitution and integrate
Substitute \( t \) and \( x dx \) back into the integral: \[ I = \int \frac{3a}{t} \cdot \left( \frac{1}{2c^2} \right) dt \] Take the constants outside the integral: \[ I = \frac{3a}{2c^2} \int \frac{1}{t} dt \] \[ I = \frac{3a}{2c^2} \log |t| + K \]

Step 3: Substitute back the original variable and compare coefficients
Replacing \( t \) with its expression in \( x \): \[ I = \frac{3a}{2c^2} \log |b^2 + c^2x^2| + K \] Comparing this result with the given form \( A \log |b^2 + c^2x^2| + K \): \[ A = \frac{3a}{2c^2} \]