\(\int\displaystyle\frac{dx}{\sqrt{3x-2x^2}}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{4x-3}{3}\right)+C\)
Taking out the coefficient 2 and completing the square:
\( \displaystyle 3x-2x^2 = 2\left(\frac{3}{2}x-x^2\right) \)
\( \displaystyle = 2\left[\frac{9}{16}-\left(x-\frac{3}{4}\right)^2\right] \)
So
\( \displaystyle \int\frac{dx}{\sqrt{3x-2x^2}} = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\left(\frac{3}{4}\right)^2-\left(x-\frac{3}{4}\right)^2}} \)
By the standard integral \(\int\frac{dt}{\sqrt{a^2-t^2}} = \sin^{-1}\frac{t}{a}+C\) with \(a = \frac{3}{4}\):
\( \displaystyle = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{x-\frac{3}{4}}{\frac{3}{4}}\right)+C \)
\( \displaystyle = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{4x-3}{3}\right)+C \)
Why the other options are incorrect
Option (1) \(\displaystyle\sqrt{2}\,\sin^{-1}\left(\frac{4x-3}{3}\right)+C\) — Incorrect. The factor \(\sqrt{2}\) comes out of the denominator, so it divides; it does not multiply.
Option (2) \(\displaystyle\frac{1}{2}\sin^{-1}\left(\frac{4x-3}{3}\right)+C\) — Incorrect. Only \(\sqrt{2}\) comes out of the square root, not 2.
Option (4) \(\displaystyle\sin^{-1}\left(\frac{4x-3}{3}\right)+C\) — Incorrect. It drops the factor \(\frac{1}{\sqrt{2}}\) left after taking 2 out of the quadratic.