The value of \(\int_{0}^{1}\displaystyle\frac{dx}{\sqrt{x+1}+\sqrt{x}}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{4}{3}\left(\sqrt{2}-1\right)\)
Multiplying the numerator and the denominator by \(\sqrt{x+1}-\sqrt{x}\), the denominator becomes \((x+1) - x = 1\):
\( \displaystyle \frac{1}{\sqrt{x+1}+\sqrt{x}} = \sqrt{x+1} - \sqrt{x} \)
So
\( \displaystyle \int_{0}^{1}\left(\sqrt{x+1} - \sqrt{x}\right)dx = \left[\frac{2}{3}(x+1)^{3/2} - \frac{2}{3}x^{3/2}\right]_{0}^{1} \)
\( \displaystyle = \frac{2}{3}\left(2\sqrt{2} - 1\right) - \frac{2}{3}(1 - 0) \)
\( \displaystyle = \frac{2}{3}\left(2\sqrt{2} - 2\right) = \frac{4}{3}\left(\sqrt{2}-1\right) \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{2}{3}\left(\sqrt{2}-1\right)\) — Incorrect. It is half the value; \(2\sqrt{2} - 2 = 2(\sqrt{2}-1)\), so the factor is \(\frac{4}{3}\).
Option (3) \(\displaystyle\frac{4}{3}\left(\sqrt{2}+1\right)\) — Incorrect. The part \(\int_{0}^{1}\sqrt{x}\,dx\) is subtracted, so the bracket is \(\sqrt{2}-1\).
Option (4) \(\displaystyle\frac{2}{3}\left(2\sqrt{2}-1\right)\) — Incorrect. This is only \(\int_{0}^{1}\sqrt{x+1}\,dx\); the part \(\int_{0}^{1}\sqrt{x}\,dx = \frac{2}{3}\) is left out.