Question

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

Answer 1

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