The integral of the function \(\displaystyle\frac{1}{9-4x^2}\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{12}\log_e\left|\frac{3+2x}{3-2x}\right|+C\), where C is an arbitrary constant
The denominator is a difference of squares, \(3^2 - (2x)^2\). Use the standard integral
\( \displaystyle \int\frac{du}{a^2-u^2} = \frac{1}{2a}\log_e\left|\frac{a+u}{a-u}\right| + C \)
Put u = 2x, so du = 2 dx, and a = 3:
\( \displaystyle \int\frac{dx}{9-4x^2} = \frac{1}{2}\int\frac{du}{3^2-u^2} \)
\( \displaystyle = \frac{1}{2}\times\frac{1}{2\times 3}\log_e\left|\frac{3+u}{3-u}\right| + C \)
\( \displaystyle = \frac{1}{12}\log_e\left|\frac{3+2x}{3-2x}\right| + C \)
Why the other options are incorrect
Differentiating each option shows which function it is the integral of.
Option (1) \(\frac{1}{22}\log_e\left|\frac{3+x}{3-x}\right|+C\), where C is an arbitrary constant — Incorrect. Its derivative is \(\frac{1}{22}\cdot\frac{6}{9-x^2} = \frac{3}{11(9-x^2)}\), not \(\frac{1}{9-4x^2}\).
Option (3) \(\frac{1}{2}\log_e\left|\frac{7+x}{7-x}\right|+C\), where C is an arbitrary constant — Incorrect. Its derivative is \(\frac{1}{2}\cdot\frac{14}{49-x^2} = \frac{7}{49-x^2}\).
Option (4) \(\frac{1}{12}\log_e\left|\frac{3-2x}{3+2x}\right|+C\), where C is an arbitrary constant — Incorrect. The fraction inside the logarithm is upside down, so this is the negative of the answer; its derivative is \(-\frac{1}{9-4x^2}\). This form belongs to \(\int\frac{dx}{4x^2-9}\).