Question

The integral of the function \( \frac{1}{9 - 4x^2} \) is:

• A. \( \frac{1}{12} \log_e \left| \frac{3 + 2x}{3 - 2x} \right| + C \), where \( C \) is an arbitrary constant
• B. \( \frac{1}{22} \log_e \left| \frac{3 + x}{3 - x} \right| + C \), where \( C \) is an arbitrary constant
• C. \( \frac{1}{2} \log_e \left| \frac{7 + x}{7 - x} \right| + C \), where \( C \) is an arbitrary constant
• D. \( \frac{1}{12} \log_e \left| \frac{3 - 2x}{3 + 2x} \right| + C \), where \( C \) is an arbitrary constant

Answer 1

The Correct Option is A

To integrate \( \frac{1}{9-4x^2} \), we factor the denominator as \(9-4x^2 = (3-2x)(3+2x)\).
The expression can be integrated using the standard form \(\int \frac{1}{a^2-x^2}dx = \frac{1}{2a}\log_e \left| \frac{a+x}{a-x} \right| + C\).
For \( \int \frac{1}{9-4x^2}dx \), we rewrite it as \( \int \frac{1}{9-(2x)^2}dx \).
Let \( u = 2x \), so \( du = 2dx \) or \( dx = \frac{1}{2}du \). Then \( \int \frac{1}{9-u^2}\left(\frac{1}{2}du\right) = \frac{1}{2}\int \frac{1}{3^2-u^2}du \).
Applying the standard form with \( a=3 \), we get \( \frac{1}{2} \left( \frac{1}{2\cdot 3}\log_e \left| \frac{3+u}{3-u} \right| \right) + C \).
Substituting back \( u=2x \), the integral is \( \frac{1}{12}\log_e \left| \frac{3+2x}{3-2x} \right| + C \).