Question

Evaluate the definite integral: $\int_{0}^{1}\frac{dx}{\sqrt{1+x}-\sqrt{x}}$

Answer 1

Let\space I=$$\int_{0}^{1}\frac{dx}{\sqrt{1+x}-\sqrt{x}}

I=\int_{0}^{1}$$\frac{1}{(\sqrt{1+x}-\sqrt{x})}\times\frac{(\sqrt{1+x}+\sqrt{x})}{(\sqrt{1+x}+\sqrt{x})}dx

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

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

=[\frac{2}{3}(1+x)^{\frac{3}{2}}]_{0}^{1}$$+[\frac{2}{3}(x)^{\frac{3}{2}}]_{0}^{1}

$=\frac{2}{3}[(2)^{\frac{2}{3}}-1]+\frac{2}{3}[1]$

$=\frac{1}{3}(2)^{\frac{3}{2}}$

$=\frac{2.2\sqrt{2}}{3}$

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