Question

Evaluate the following definite integrals: $\int^1_0 {\frac{dx}{\sqrt{1+x}-\sqrt{x}}$

Answer 1

Answer:

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

Step-by-step explanation:

$Formula\:used:\\\\\int{x^n}\:dx=\frac{x^{n+1}}{n+1}+c$

$\int\limits^1_{0}\frac{1}{\sqrt{1+x}+\sqrt{x}}\:dx$

$=\int\limits^1_0\frac{1}{\sqrt{1+x}-\sqrt{x}}.\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}\:dx$

$=\int\limits^1_0\frac{\sqrt{1+x}+\sqrt{x}}{1+x-x}\:dx$

$=\int\limits^1_0[\sqrt{1+x}+\sqrt{x}}]\:dx\\\\=\int\limits^1_0[(1+x)^{\frac{1}{2}}+x^{\frac{1}{2}}]\:dx$

$=[\frac{(1+x)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]^1_0\\\\=\frac{2}{3}[(1+x)^{\frac{3}{2}}+x^{\frac{3}{2}}]^1_0$

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

$=\frac{2}{3}[2^{\frac{3}{2}}+1]-\frac{2}{3}[1]\\\\=\frac{2}{3}2^{\frac{3}{2}}+\frac{2}{3}-\frac{2}{3}\\\\=\frac{2^{\frac{5}{2}}}{3}\\\\=\frac{4\sqrt{2}}{3}$