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

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

Answers (1)

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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}

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