Question

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

Answer 1

$$\int \frac{dx}{(\sqrt{x+a}+\sqrt{x+b})}$$
$$=\int \frac{(\sqrt{x+a}-\sqrt{x-b})}{(\sqrt{x+a}+\sqrt{x+b})(\sqrt{x+a}-\sqrt{x+b})}dx$$
$$=\int \frac{(\sqrt{x+a}-\sqrt{x+b})}{(x+a)-(x+b)}dx$$
$$=\frac{1}{a-b}\int {(x+a)}^{\frac{1}{2}}-\frac{1}{a-b}\int {(x+b)}^{\frac{1}{2}}dx$$
${\displaystyle =\frac{1}{(a-b)}\left[\frac{{(x+a)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}$ - ${\displaystyle \frac{1}{a-b}\left[\frac{{(x+b)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}$
$$=\frac{2}{3(a-b)}[{(x+a)}^{\frac{3}{2}}-{(x+b)}^{\frac{3}{2}}]+C$$