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: Pull a constant outside.
Take $\dfrac{1}{4}$ out so the $x^2$ stands alone.
\[ \int \frac{dx}{9 - 4x^2} = \frac{1}{4}\int \frac{dx}{\frac{9}{4} - x^2} \]

Step 2: Spot the value of $a$.
Here $\dfrac{9}{4} = \left(\dfrac{3}{2}\right)^2$, so $a = \dfrac{3}{2}$.

Step 3: Use the standard form.
We know $\displaystyle\int \frac{dx}{a^2 - x^2} = \frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|$. With $a = \frac{3}{2}$ the factor is $\dfrac{1}{2a} = \dfrac{1}{3}$.
\[ \frac{1}{4}\cdot \frac{1}{3}\log\left|\frac{\frac{3}{2}+x}{\frac{3}{2}-x}\right| + C \]

Step 4: Clean the fraction inside.
Multiply top and bottom inside the log by $2$. The outside becomes $\dfrac{1}{12}$.
\[ \frac{1}{12}\log\left|\frac{3+2x}{3-2x}\right| + C \]
That is option 2.
\[ \boxed{\dfrac{1}{12}\log_e\left|\dfrac{3+2x}{3-2x}\right| + C} \]