Question

Solve: $\stackrel{1}{\underset{0}{\int }}\frac{dx}{\sqrt{x+1}+\sqrt{x}}$

Answer 1

${\int }_0^1\frac{dx}{\sqrt{x+1}+\sqrt{x}}$

${\int }_0^1\frac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1}-\sqrt{x})}\frac{dx}{(\sqrt{x+1}+\sqrt{x})}$

${\int }_0^1\frac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1}{)}^2-(\sqrt{x}{)}^2}dx$

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

${\int }_0^1\sqrt{x+1}-\sqrt{x}dx$

$2\left(\frac{(x+1{)}^{\frac{3}{2}}}{3}\right)-\frac{2x^{\frac{3}{2}}}{3}{|}_0^1$

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

$\therefore {\int }_0^1\frac{dx}{\sqrt{x+1}+\sqrt{x}}=\frac{2}{3}(\frac{2^{\frac{3}{2}}}{3}-1)$