\(\int\displaystyle\frac{x + 2}{\sqrt{3x + 1}}\,dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{2}{27}(3x + 1)^{3/2} + \frac{10}{9}\sqrt{3x + 1} + C\)
Let \(t = 3x + 1\). Then
\( \displaystyle x = \frac{t - 1}{3} \)
\( \displaystyle dx = \frac{1}{3}\,dt \)
The numerator in terms of t:
\( \displaystyle x + 2 = \frac{t - 1}{3} + 2 = \frac{t + 5}{3} \)
So the integral becomes
\( \displaystyle \int \frac{t + 5}{3\sqrt{t}} \cdot \frac{1}{3}\,dt = \frac{1}{9}\int \left(t^{1/2} + 5t^{-1/2}\right)dt \)
\( \displaystyle = \frac{1}{9}\left(\frac{2}{3}t^{3/2} + 10t^{1/2}\right) + C \)
\( \displaystyle = \frac{2}{27}(3x + 1)^{3/2} + \frac{10}{9}\sqrt{3x + 1} + C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{2}{9}(3x + 1)^{3/2} + \frac{10}{3}\sqrt{3x + 1} + C\) — Incorrect. This is three times the answer: the factor \(\frac{1}{3}\) from \(dx = \frac{1}{3}\,dt\) is left out.
Option (2) \(\displaystyle\frac{2}{27}(3x + 1)^{3/2} - \frac{10}{9}\sqrt{3x + 1} + C\) — Incorrect. The second term has the wrong sign; \(x + 2 = \frac{t + 5}{3}\), not \(\frac{t - 5}{3}\).
Option (3) \(\displaystyle\frac{1}{9}(3x + 1)^{3/2} + \frac{5}{3}\sqrt{3x + 1} + C\) — Incorrect. Its derivative at \(x = 0\) is \(\frac{1}{2} + \frac{5}{2} = 3\), but the integrand at \(x = 0\) is 2.