\(\int\sqrt{x}\,\log x\,dx\) is equal to (\(x \gt 0\), C is a constant of integration)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{2}{9}x^{3/2}\left(3\log x-2\right)+C\)
Using integration by parts with \(\log x\) as the first function and \(x^{1/2}\) as the second:
\( \displaystyle \int x^{1/2}\,dx = \frac{2}{3}x^{3/2} \)
\( \displaystyle \int\sqrt{x}\,\log x\,dx = \frac{2}{3}x^{3/2}\log x-\int\frac{2}{3}x^{3/2}\cdot\frac{1}{x}\,dx \)
\( \displaystyle = \frac{2}{3}x^{3/2}\log x-\frac{2}{3}\int x^{1/2}\,dx \)
\( \displaystyle = \frac{2}{3}x^{3/2}\log x-\frac{4}{9}x^{3/2}+C \)
Taking \(\frac{2}{9}x^{3/2}\) common:
\( \displaystyle = \frac{2}{9}x^{3/2}\left(3\log x-2\right)+C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{2}{9}x^{3/2}\left(3\log x+2\right)+C\) — Incorrect. The second term of by parts is subtracted, so it is \(-\frac{4}{9}x^{3/2}\), not \(+\frac{4}{9}x^{3/2}\).
Option (3) \(\displaystyle\frac{2}{3}x^{3/2}\left(\log x-1\right)+C\) — Incorrect. It forgets to integrate \(\frac{2}{3}x^{1/2}\) again; that gives \(\frac{4}{9}x^{3/2}\), not \(\frac{2}{3}x^{3/2}\).
Option (4) \(\displaystyle\frac{2}{9}x^{3/2}\left(2\log x-3\right)+C\) — Incorrect. The coefficient of \(x^{3/2}\log x\) must be \(\frac{2}{3} = \frac{6}{9}\), but here it is \(\frac{4}{9}\).