Question

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

Answer 1

$$\begin{gathered} \int \frac{1}{\sqrt{x+1}+\sqrt{x}}dx \\ \mathrm{Rationalising}\mathrm{the}\mathrm{denominator}, \\ =\int \frac{\left(\sqrt{x+1}-\sqrt{x}\right)}{\left(\sqrt{x+1}+\sqrt{x}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}dx \\ =\int \frac{\left(\sqrt{x+1}-\sqrt{x}\right)}{x+1-x}dx \\ =\int \left(\sqrt{x+1}-\sqrt{x}\right)dx \\ =\int \left[{\left(x+1\right)}^{\frac{1}{2}}-x^{\frac{1}{2}}\right]dx \\ =\frac{{\left(x+1\right)}^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}-\frac{x^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}+C \\ =\frac{2}{3}{\left(x+1\right)}^{\frac{3}{2}}-\frac{2}{3}{x}^{\frac{3}{2}}+C \\ \mathit{=}\frac{\mathit{2}}{\mathit{3}}\left[{\left(\mathit{x}\mathit{+}\mathit{1}\right)}^{\frac{\mathit{3}}{\mathit{2}}}\mathit{-}\mathit{}{\mathit{x}}^{\frac{\mathit{3}}{\mathit{2}}}\right]+C \end{gathered}$$