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

Begin by rewriting the denominator \(9 - 4x^2\):

\[ 9 - 4x^2 = (3 - 2x)(3 + 2x). \]

The integral becomes:

\[ \int \frac{1}{9 - 4x^2} dx = \int \frac{1}{(3 - 2x)(3 + 2x)} dx. \]

Using partial fraction decomposition:

\[ \frac{1}{(3 - 2x)(3 + 2x)} = \frac{A}{3 - 2x} + \frac{B}{3 + 2x}. \]

Equating and solving for \(A\) and \(B\):

\[ A = \frac{1}{6}, \quad B = \frac{1}{6}. \]

The integral becomes:

\[ \int \frac{1}{9 - 4x^2} dx = \frac{1}{6} \int \frac{1}{3 - 2x} dx + \frac{1}{6} \int \frac{1}{3 + 2x} dx. \]

Solving each term:

\[ \int \frac{1}{3 - 2x} dx = -\frac{1}{2} \log_e |3 - 2x|, \quad \int \frac{1}{3 + 2x} dx = \frac{1}{2} \log_e |3 + 2x|. \]

Substituting back:

\[ \int \frac{1}{9 - 4x^2} dx = \frac{1}{6} \left(-\frac{1}{2} \log_e |3 - 2x| + \frac{1}{2} \log_e |3 + 2x| \right). \]

Simplify:

\[ \int \frac{1}{9 - 4x^2} dx = \frac{1}{12} \log_e \left|\frac{3 + 2x}{3 - 2x}\right| + C. \]