Answer
$$\int \frac{1}{\sqrt{x+1}+\sqrt{x}}dx=\frac{2}{3}\left [ (x+1)^{\frac{3}{2}}-x^{\frac{3}{2}} \right ]+C$$
Work Step by Step
$$\int \frac{1}{\sqrt{x+1}+\sqrt{x}}dx$$
$$=\int \frac{1}{\sqrt{x+1}+\sqrt{x}}\cdot \frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}dx$$
$$=\int (\sqrt{x+1}-\sqrt{x})dx=\frac{2}{3}\left [ (x+1)^{\frac{3}{2}}-x^{\frac{3}{2}} \right ]+C$$
