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

Answer: (A)

$\frac{1}{12} \log_e \left| \frac{3 + 2x}{3 - 2x} \right| + C$, where $C$ is an arbitrary constant

Answer 2

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:

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