$\int\limits_0^1\frac{dx}{\sqrt{1+x}-\sqrt{x}}$ is equal to

$\frac{4\sqrt{2}}{3}$

The correct answer is Option (2) → $\frac{4\sqrt{2}}{3}$

Evaluate the integral:

$\int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}$

Rationalize the denominator:

$\frac{1}{\sqrt{1+x}-\sqrt{x}} \cdot \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}} =\frac{\sqrt{1+x}+\sqrt{x}}{(1+x)-x} =\sqrt{1+x}+\sqrt{x}$

Thus the integral becomes:

$\int_{0}^{1} \left(\sqrt{1+x}+\sqrt{x}\right)\,dx$

$\int_{0}^{1}\sqrt{1+x}\,dx+\int_{0}^{1}\sqrt{x}\,dx$

Compute each:

$\int\sqrt{1+x}\,dx=\frac{2}{3}(1+x)^{3/2}$

$\int\sqrt{x}\,dx=\frac{2}{3}x^{3/2}$

$\frac{2}{3}(2\sqrt{2}-1)+\frac{2}{3} =\frac{2}{3}(2\sqrt{2}-1+1) =\frac{2}{3}(2\sqrt{2})$

$=\frac{4\sqrt{2}}{3}$

The value of the integral is $\frac{4\sqrt{2}}{3}$.