If \(\int\displaystyle\frac{dx}{\sqrt{x+2}-\sqrt{x+1}}=\displaystyle\frac{2}{3}\left[(\lambda+1)^{\frac{3}{2}}+\lambda^{\frac{3}{2}}\right]+C\), then the value of \(\lambda\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → x + 1
Rationalise the denominator by multiplying the numerator and the denominator by \(\sqrt{x+2}+\sqrt{x+1}\):
\( \displaystyle \frac{1}{\sqrt{x+2}-\sqrt{x+1}} = \frac{\sqrt{x+2}+\sqrt{x+1}}{(x+2)-(x+1)} \)
\( = \sqrt{x+2}+\sqrt{x+1} \)
So
\( \displaystyle \int\frac{dx}{\sqrt{x+2}-\sqrt{x+1}} = \int\left((x+2)^{1/2}+(x+1)^{1/2}\right)dx \)
\( \displaystyle = \frac{2}{3}\left[(x+2)^{3/2}+(x+1)^{3/2}\right]+C \)
Comparing with \(\frac{2}{3}\left[(\lambda+1)^{3/2}+\lambda^{3/2}\right]+C\): with \(\lambda = x + 1\), \(\lambda + 1 = x + 2\), and the two expressions match term by term. So \(\lambda = x + 1\).