\(\int\displaystyle\frac{3}{\sqrt{3x - 2}}\,dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2\sqrt{3x - 2} + C\)
Let \(t = 3x - 2\). Then \(dt = 3\,dx\), so \(3\,dx = dt\).
\( \displaystyle \int \frac{3}{\sqrt{3x - 2}}\,dx = \int t^{-1/2}\,dt \)
\( \displaystyle = \frac{t^{1/2}}{1/2} + C \)
\( = 2\sqrt{3x - 2} + C \)
Check by differentiating: \(\frac{d}{dx}\left(2\sqrt{3x - 2}\right) = 2 \cdot \frac{3}{2\sqrt{3x - 2}} = \frac{3}{\sqrt{3x - 2}}\).