The value of \(\int_{0}^{1}\displaystyle\frac{x}{\sqrt{1+x}}\,dx\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{4-2\sqrt{2}}{3}\)
Let \(t = 1+x\), so \(dt = dx\) and \(x = t-1\).
When \(x = 0\), \(t = 1\); when \(x = 1\), \(t = 2\). Then
\( \displaystyle \int_{0}^{1}\frac{x}{\sqrt{1+x}}\,dx = \int_{1}^{2}\frac{t-1}{\sqrt{t}}\,dt \)
\( \displaystyle = \int_{1}^{2}\left(t^{1/2} - t^{-1/2}\right)dt \)
\( \displaystyle = \left[\frac{2}{3}t^{3/2} - 2t^{1/2}\right]_{1}^{2} \)
\( \displaystyle = \left(\frac{4\sqrt{2}}{3} - 2\sqrt{2}\right) - \left(\frac{2}{3} - 2\right) \)
\( \displaystyle = -\frac{2\sqrt{2}}{3} + \frac{4}{3} = \frac{4-2\sqrt{2}}{3} \)