The given function is \( \displaystyle \frac{1}{\sqrt{x + a} + \sqrt{x + b}} \) and we need to find the integration of this function with respect to \( x \).
Solution:
\( \begin{aligned} &\int \frac{dx}{\sqrt{x + a} + \sqrt{x + b}} \\ &= \int \dfrac{\sqrt{x+a}-\sqrt{x+b}}{(\sqrt{x+a}+\sqrt{x+b})(\sqrt{x+a}-\sqrt{x+b})} dx \\ &= \int \dfrac{\sqrt{x+a}-\sqrt{x+b}}{(\sqrt{x+a})^2-(\sqrt{x+b})^2}dx \quad [\because (a+b)(a-b)=a^2-b^2] \\ &= \int \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} dx \\ &= \frac{1}{a-b} \int \left(\sqrt{x+a}-\sqrt{x+b} \right) dx \\ &= \frac{1}{a-b} \left( \int \sqrt{x+a} \, dx-\int \sqrt{x+b} \,dx \right) \\ &= \frac{1}{a-b} \left[ \frac{2}{3}(x+a)^{3/2}-\frac{2}{3}(x+b)^{3/2} \right] + C, \quad \text{(C is a constant)} \\ &= \frac{2}{3(a-b)} \left[ (x+a)^{3/2}-(x+b)^{3/2} \right] + C \\ \end{aligned} \)Thus, the integration of the given function is
\( \boxed{\int \frac{dx}{\sqrt{x + a} + \sqrt{x + b}} = \frac{2}{3(a – b)} \left[ (x + a)^{3/2} – (x + b)^{3/2} \right] + C } \)Please let me know in the comments if you find any errors in this solution.