Question

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

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

Hint:

To avoid confusion between \( a^2-x^2 \) and \( x^2-a^2 \), remember:
If \( x \) is subtracted (\( a^2-x^2 \)), the log has \( (a+x) \) in the numerator.
If \( a \) is subtracted (\( x^2-a^2 \)), the log has \( (x-a) \) in the numerator.
Check the coefficient of \( x^2 \) first! If it's 4, your multiplier will be \( \frac{1}{\sqrt{4}} \times \frac{1}{2a} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves indefinite integration, specifically of a rational function where the denominator is a quadratic expression of the form \( a^2 - x^2 \). This category of integrals is standard in calculus and is solved using partial fraction decomposition or by applying a specialized formula for inverse hyperbolic/logarithmic forms.
The expression \( \frac{1}{a^2 - x^2} \) integrates to a logarithmic function because its partial fraction components are of the form \( \frac{1}{a-x} + \frac{1}{a+x} \), and the integral of \( \frac{1}{x} \) is \( \log|x| \). A key detail in these problems is handling the coefficients of \( x^2 \) correctly to ensure the final scaling factor is accurate.

Step 2: Key Formula or Approach:
We can use the general integration formula:
\[ \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C \]
Alternatively, we can use the method of substitution if the \( x^2 \) term has a coefficient. For an integral of the form \( \int \frac{dx}{a^2 - k^2x^2} \), we substitute \( u = kx \), which results in a factor of \( \frac{1}{k} \) being pulled out front. Combining this with the formula gives:
\[ \int \frac{dx}{a^2 - (kx)^2} = \frac{1}{k} \cdot \frac{1}{2a} \log \left| \frac{a+kx}{a-kx} \right| + C \]

Step 3: Detailed Explanation:
Step 3.1: Identifying parameters.
The given expression is \( \frac{1}{9 - 4x^2} \).
We can rewrite this as:
\[ \frac{1}{3^2 - (2x)^2} \]
Here, \( a = 3 \) and the term being squared is \( 2x \).
Step 3.2: Applying substitution.
Let \( 2x = t \).
Differentiating both sides: \( 2 dx = dt \implies dx = \frac{dt}{2} \).
Substituting these into the integral:
\[ I = \int \frac{1}{3^2 - t^2} \cdot \frac{dt}{2} = \frac{1}{2} \int \frac{dt}{3^2 - t^2} \]
Step 3.3: Evaluating using the standard formula.
Using \( \int \frac{dt}{a^2 - t^2} = \frac{1}{2a} \log \left| \frac{a+t}{a-t} \right| \) with \( a = 3 \):
\[ I = \frac{1}{2} \left[ \frac{1}{2(3)} \log \left| \frac{3+t}{3-t} \right| \right] + C \]
\[ I = \frac{1}{2} \cdot \frac{1}{6} \log \left| \frac{3+t}{3-t} \right| + C \]
Step 3.4: Re-substituting and simplifying.
Substitute \( t = 2x \) back into the expression:
\[ I = \frac{1}{12} \log \left| \frac{3+2x}{3-2x} \right| + C \]
Comparing this result with the provided options, we see it perfectly matches Option (2). Note that Option (4) is a distractor with the numerator and denominator swapped; the formula for \( a^2 - x^2 \) always has \( a+x \) on top.

Step 4: Final Answer:
By utilizing the standard integration form for the difference of squares and carefully managing the coefficient of \( x \), we obtain \( \frac{1}{12} \log_e \left| \frac{3+2x}{3-2x} \right| + C \). Thus, Option (2) is correct.