Question

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

Answer 1

$\frac{1}{\sqrt{x+a}+\sqrt{x+b}}$=$\frac{1}{\sqrt{x+a}+\sqrt{x+b}}$×$\sqrt{x+a}$-$\frac{\sqrt{x+b}}{\sqrt{x+a}}$-$\sqrt{x+b}$

=$\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}$

⇒$\int \frac{1}{\sqrt{x+a}}=\sqrt{x}+bdx=\frac{1}{a}-b\int (\sqrt{x+a}=\sqrt{x+b})dx$

=$\frac{2}{3}$(a-b)[(x+a)$^{\frac{3}{2}}$-(x+b)$^{\frac{3}{2}}$]+C