\(\int\displaystyle\frac{dx}{4x^2-9}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{12}\log\left|\frac{2x-3}{2x+3}\right|+C\)
Let \(t = 2x\), so \(dx = \frac{dt}{2}\) and \(4x^2-9 = t^2-3^2\):
\( \displaystyle \int\frac{dx}{4x^2-9} = \frac{1}{2}\int\frac{dt}{t^2-3^2} \)
Using the standard integral \(\int\frac{dt}{t^2-a^2} = \frac{1}{2a}\log\left|\frac{t-a}{t+a}\right|\) with a = 3:
\( \displaystyle = \frac{1}{2}\times\frac{1}{6}\log\left|\frac{2x-3}{2x+3}\right|+C \)
\( \displaystyle = \frac{1}{12}\log\left|\frac{2x-3}{2x+3}\right|+C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{6}\log\left|\frac{2x-3}{2x+3}\right|+C\) — Incorrect. It drops the factor \(\frac{1}{2}\) that comes from \(dx = \frac{dt}{2}\).
Option (2) \(\displaystyle\frac{1}{12}\log\left|\frac{2x+3}{2x-3}\right|+C\) — Incorrect. The fraction is upside down; this is the result for \(\frac{1}{9-4x^2}\).
Option (4) \(\displaystyle\frac{1}{3}\log\left|\frac{2x-3}{2x+3}\right|+C\) — Incorrect. The factor in front is \(\frac{1}{2}\times\frac{1}{2a} = \frac{1}{12}\), not \(\frac{1}{3}\).