Question

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

Answer 1

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

Rationalise the denominator

$$\begin{gathered} =\int \frac{\left(\sqrt{x+1}-\sqrt{x}\right)}{\left(\sqrt{x+1}+\sqrt{x}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}dx \\ =\int \frac{\left(\sqrt{x+1}-\sqrt{x}\right)}{\left(x+1\right)-x}dx \\ =\int {\left(x+1\right)}^{\frac{1}{2}}dx-\int x^{\frac{1}{2}}dx \\ =\frac{{\left(x+1\right)}^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}-\frac{x^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1} \\ =\frac{2}{3}{\left(x+1\right)}^{\frac{3}{2}}-\frac{2}{3}{x}^{\frac{3}{2}}+\mathrm{C} \end{gathered}$$