Integrate the function: $\frac{1}{\sqrt{x+a}+\sqrt{x+b}}$

To integrate $\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} dx$,we first rationalize the denominator:
$\frac{1}{\sqrt{x+a}+\sqrt{x+b}} = \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}$
$= \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} = \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b}$
Now,integrate the expression:
$\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} dx = \frac{1}{a-b} \int (\sqrt{x+a}-\sqrt{x+b}) dx$
$= \frac{1}{a-b} \left[ \int (x+a)^{\frac{1}{2}} dx - \int (x+b)^{\frac{1}{2}} dx \right]$
$= \frac{1}{a-b} \left[ \frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}} - \frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}} \right] + C$
$= \frac{2}{3(a-b)} \left[ (x+a)^{\frac{3}{2}} - (x+b)^{\frac{3}{2}} \right] + C$