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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 3

Answers (1)

Answer : -        2 \sqrt{x+2}+\frac{2}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C

Hint :-                Use substitution method to solve this integral.

Given :-              \int \frac{x+1}{(x-1) \sqrt{x+2}} d x \\

Sol : -  

\begin{aligned} &\text { Let } \mathrm{I}=\int \frac{x+1}{(x-1) \sqrt{x+2}} d x \\ &=\int \frac{x-1+2}{(x-1) \sqrt{x+2}} d x \\ &\Rightarrow \mathrm{I}=\int \frac{(x-1)+2}{(x-1) \sqrt{x+2}} d x \\ &\Rightarrow \mathrm{I}=\int\left\{\frac{(x-1)}{(x-1) \sqrt{x+2}}+\frac{2}{(x-1) \sqrt{x+2}}\right\} d x \end{aligned}

\Rightarrow \mathrm{I}=\int \frac{1}{\sqrt{x+2}} d x+\int \frac{2}{(x-1) \sqrt{x+2}} d x  

Now, I = I1 +I2     ........................ (I)

            Where \mathrm{I}_{1}=\int \frac{1}{\sqrt{x+2}} d x \text { and } \mathrm{I}_{2}=\int \frac{2}{(x-1) \sqrt{x+2}}dx

 Therefore, I_{1}=\int \frac{1}{\sqrt{x+2}} d x

                     Let \mathrm{x}+2=\mathrm{u} \Rightarrow \mathrm{d} \mathrm{x}=\mathrm{du} Then,

                \begin{aligned} \mathrm{I}_{1} &=\int \frac{1}{u^{\frac{1}{2}}} d u=\int u^{\frac{-1}{2}} \mathrm{du}=\frac{u^{\frac{-1}{2}}+1}{\frac{-1}{2}+1}+C \quad\left(\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right) \\ &=\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+\mathrm{C}_{1} \\ &=2 \sqrt{x+2}+\mathrm{C}_{1} \ldots \ldots \ldots \ldots \text { (II) } \end{aligned}

 And I_{2}=\int \frac{2}{(x-1) \sqrt{x+2}} d x

Let x+2=t^{2}=>d x=2 t d t. Then,

\begin{aligned} \mathrm{I}_{2} &=\int \frac{2}{\left(t^{2}-2-1\right) \sqrt{t^{2}}} 2 \mathrm{t} \mathrm{d} \mathrm{t}=4 \int \frac{t d t}{\left(t^{2}-3\right) t} & &\left(\because \mathrm{t}^{2}-2=\mathrm{x}\right) \\ &=4 \int \frac{1}{\left(t^{2}\right)-\sqrt{3^{2}}} \mathrm{dt}=\frac{4}{2 \cdot \sqrt{3}} \log \left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|+\mathrm{C} & &\left(\because \int \frac{1}{x^{2}-a^{2}} \mathrm{~d} \mathrm{x}=1 / 2 \mathrm{a} \log \left|\frac{x-a}{x+a}\right|+\mathrm{C}\right) \\ &=\frac{2}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+\mathrm{C}_{2} \quad &(\because \mathrm{t}=\sqrt{x+2}) \ldots \ldots \ldots . \mathrm{III}) \end{aligned}


Put the value of Equ.(II) and (III) in (I) then,

\begin{aligned} &\mathrm{I}=2 \sqrt{x+2}+\mathrm{C}_{1}+\frac{2}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+\mathrm{C}_{2} \\ &\quad=2 \sqrt{x+2}++\frac{2}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C \quad\left(\because \mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}\right) \text { Ans.. } \end{aligned}


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