Question

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

Answer 1

Solution:

Rationalised the denominator

$\frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a} - \sqrt{x+b}}{\sqrt{x+a} - \sqrt{x+b}} \\ \\ = \frac{\sqrt{x+a} - \sqrt{x+b}}{x + a - x - b} \\ \\ = \int\frac{\sqrt{x+a} - \sqrt{x+b}}{a - b} dx \\ \\ = \int\frac{1}{a - b} \sqrt{x + a} \: dx - \int\frac{1}{a - b} \sqrt{x + b} \: dx \\ \\ = \frac{2}{3(a - b)} {(x + a)}^{ \frac{3}{2} } - \frac{2}{3(a - b)} {(x + b)}^{ \frac{3}{2} } + C \\ \\\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}}= \frac{2}{3(a - b)} \bigg({(x + a)}^{ \frac{3}{2} } - {(x + b)}^{ \frac{3}{2} }\bigg) + C$
Hope it helps you.