Question

If \(\int _0^1\frac{dx}{\sqrt{x+1}-\sqrt{x}} = \sqrt{2}k\), then the value of \(k\) is...

• A. \(\frac{4}{\sqrt{3}}\)
• B. \(\frac{4}{3}\)
• C. \(\frac{2}{\sqrt{3}}\)
• D. \(\frac{2}{3}\)

Hint:

Multiply by the conjugate: 1/(sqrt(x+1) - sqrt x) = sqrt(x+1) + sqrt x.

Answer 1

The Correct Option is B

Step 1: Rationalise:
\[ \frac{1}{\sqrt{x+1} - \sqrt{x}} \cdot \frac{\sqrt{x+1} + \sqrt{x}}{\sqrt{x+1} + \sqrt{x}} = \frac{\sqrt{x+1} + \sqrt{x}}{(x+1) - x} = \sqrt{x+1} + \sqrt{x} \]

Step 2: Integrate:
\[ \int_0^1 (\sqrt{x+1} + \sqrt{x})dx = \frac23\left[(x+1)^{3/2} + x^{3/2}\right]_0^1 \]

Step 3: Evaluate:
At \(x = 1\): \(2^{3/2} + 1 = 2\sqrt2 + 1\). At \(x = 0\): \(1 + 0 = 1\).
\[ \frac23\left[(2\sqrt2 + 1) - 1\right] = \frac23 \cdot 2\sqrt2 = \frac{4\sqrt2}{3} \]
Comparing with \(\sqrt2\,k\) gives \(k = \frac43\).

Step 4: Why the other options are wrong.
\(\frac{4}{\sqrt3}\) and \(\frac2{\sqrt3}\) contain a root of 3, which never appears here. \(\frac23\) forgets the factor 2 in \(2\sqrt2\).

Final Answer:
\(k = \frac43\), option (B). \[ \boxed{\frac{4}{3}} \]