Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.5 Question 10 Maths Textbook Solution.

Answer: $\frac{1}{a-b}\left\{\frac{2}{5}(x+a)^{\frac{5}{2}}-\frac{2 a}{3}(x+a)^{\frac{3}{2}}+\frac{2}{5}(x+b)^{\frac{5}{2}}-\frac{2 b}{3}(x+b)^{\frac{3}{2}}\right\}+C$

Hint:  Use: $(a-b)(a b)=a^{2}-b^{2}$

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

Solution:

\begin{aligned} &=\int \frac{x}{\sqrt{x+a}-\sqrt{x+b}} \Rightarrow \frac{x}{\sqrt{x+a}-\sqrt{x+b}} \Rightarrow \frac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}} \\ &=\frac{x(\sqrt{x+a}+\sqrt{x+b})}{(\sqrt{x+a})^{2}-(\sqrt{x+b})^{2}} \Rightarrow \frac{x \sqrt{x+a}+x \sqrt{x+b}}{x+a-(x+b)} \end{aligned}
\begin{aligned} &\Rightarrow \frac{1}{a-b}\left\{(x+a)^{\frac{3}{2}}-a(x+a)^{\frac{1}{2}}+(x+b)^{\frac{3}{2}}-b(x+b)^{\frac{1}{2}}\right\} \\ &\Rightarrow \int \frac{1}{\sqrt{x+a}-\sqrt{x+b}} d x=\int \frac{1}{(a-b)}\left\{(x+a)^{\frac{3}{2}}-a(x+a)^{\frac{1}{2}}+(x+b)^{\frac{3}{2}}-b(x+b)^{\frac{1}{2}}\right\} \end{aligned}

\begin{aligned} &\Rightarrow \int \frac{1}{\sqrt{x+a}-\sqrt{x+b}} d x=\int \frac{1}{(a-b)}\left\{(x+a)^{\frac{3}{2}}-a(x+a)^{\frac{1}{2}}+(x+b)^{\frac{3}{2}}-b(x+b)^{\frac{1}{2}}\right\} \\ &\Rightarrow \frac{1}{a-b}\left\{(x+a)^{\frac{3}{2}} d x-\int a(x+a)^{\frac{1}{2}} d x+\int(x+b)^{\frac{3}{2}} d x-\int b(x+b)^{\frac{1}{2}} d x\right\} \end{aligned}

\begin{aligned} &\Rightarrow \frac{1}{a-b}\left\{\frac{1}{1+\frac{3}{2}}(x+a)^{\frac{3}{2}}-a \frac{1}{1+\frac{1}{2}}(x+a)^{\frac{1}{2}}+\frac{1}{1+\frac{3}{2}}(x+b)^{\frac{3}{2}}-b \frac{1}{1+\frac{1}{2}}(x+b)^{\frac{1}{2}}\right\} \\ &=\frac{1}{a-b}\left\{\frac{2}{5}(x+a)^{\frac{5}{2}}-\frac{2 a}{3}(x+a)^{\frac{3}{2}}+\frac{2}{5}(x+b)^{\frac{5}{2}}-\frac{2 b}{3}(x+b)^{\frac{3}{2}}\right\}+C \end{aligned}