#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.3 Question 8 Maths Textbook Solution.

Answer: $\frac{2}{3(a-b)}\left\{(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right\}+c$

Hint:$\text { To solve this equation we add } \sqrt{x+a}-\sqrt{x+b} \text { to numerator and denominator }$

Given: $\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x$

Solution: $\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x$

$\text { Multiply } \sqrt{x+a}-\sqrt{x+b} \text { to numerator and denominator }$

\begin{aligned} &\int \frac{\sqrt{x+a}-\sqrt{x+b}}{(\sqrt{x+a}+\sqrt{x+b})(\sqrt{x+a}-\sqrt{x+b})} d x \\ &=\int \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} d x \end{aligned}

\begin{aligned} &=\int \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b} d x \\ &=\frac{1}{a-b} \int \sqrt{x+a}-\sqrt{x+b} d x\left[\int(x)^{n}=\frac{(x)^{n+1}}{n+1}+c\right] \\ &=\frac{1}{a-b}\left[\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+b)^{\frac{8}{2}}}{\frac{3}{2}}\right]+c \\ &=\frac{1}{a-b} \times \frac{2}{3}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{8}{2}}\right]+c \\ &=\frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{8}{2}}\right]+c \end{aligned}