\(\int\displaystyle\frac{dx}{9x^2-16}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{24}\log_e\left|\frac{3x-4}{3x+4}\right|+C\), Where C is constant of integration
Taking 9 out of the denominator brings the integral to the form \(x^2 - a^2\) with \(a = \frac{4}{3}\):
\(\displaystyle \int\frac{dx}{9x^2-16} = \frac{1}{9}\int\frac{dx}{x^2 - \left(\frac{4}{3}\right)^2}\)
By the standard integral
\(\displaystyle \int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\log_e\left|\frac{x-a}{x+a}\right| + C\)
with \(\frac{1}{2a} = \frac{3}{8}\):
\(\displaystyle \int\frac{dx}{9x^2-16} = \frac{1}{9}\cdot\frac{3}{8}\log_e\left|\frac{x-\frac{4}{3}}{x+\frac{4}{3}}\right| + C\)
Multiplying the numerator and denominator inside the modulus by 3:
\(\displaystyle = \frac{1}{24}\log_e\left|\frac{3x-4}{3x+4}\right| + C\)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{24}\log_e\left|\frac{3x+4}{3x-4}\right|+C\), Where C is constant of integration — Incorrect. The fraction is upside down. This is the negative of the answer, and it is the integral of \(\frac{1}{16 - 9x^2}\) (the \(a^2 - x^2\) form), not of \(\frac{1}{9x^2 - 16}\).
Option (2) \(\displaystyle\frac{3}{8}\log_e\left|\frac{3x+4}{3x-4}\right|+C\), Where C is constant of integration — Incorrect. The factor \(\frac{1}{9}\) is missing and the fraction is upside down.
Option (3) \(\displaystyle\frac{3}{8}\log_e\left|\frac{3x-4}{3x+4}\right|+C\), Where C is constant of integration — Incorrect. This uses \(\frac{1}{2a} = \frac{3}{8}\) but leaves out the factor 1/9 taken out of the denominator.