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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 4

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 4

Answers (1)

Answer : \frac{2}{3}(x+2)^{\frac{3}{2}}-2 \sqrt{x+2}+\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C

Hint :-   Use substitution method and special integration formula to solve this problem.

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

Sol :-

\begin{aligned} \text { let } I &=\int \frac{x^{2}}{(x-1) \sqrt{x+2}} d x \\ &=\int \frac{\left(x^{2}-1\right)+1}{(x-1) \sqrt{x+2}} d x=\int \frac{\left(x^{2}-1\right)+1}{(x-1) \sqrt{x+2}} d x \\ &=\int\left\{\frac{x^{2}-1}{(x-1) \sqrt{x+2}}+\frac{1}{(x-1) \sqrt{x+2}}\right\} d x \\ &=\int\left\{\frac{(x-1)(x+1)}{(x-1) \sqrt{x+2}}+\frac{1}{(x-1) \sqrt{x+2}}\right\} d x \quad\quad\quad\quad\left(\because a^{2}-b^{2}=(a+b)(a-b)\right) \end{aligned}

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

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

            Now,\int \sqrt{x+2}dx

            Put \mathrm{x}+2=\mathrm{p}=>\mathrm{d} \mathrm{x}=\mathrm{d} \mathrm{p} then

\begin{aligned} \int \sqrt{x+2} d x &=\int \sqrt{p} d p=\int p^{\frac{1}{2} d}=\frac{p \frac{1}{2}+1}{\frac{1}{2}+1}+\mathrm{C}_{1} & &\left(\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right) \\ &=\frac{p^{\frac{8}{2}}}{3 / 2}=2 / 3(x+2)^{3 / 2}+\mathrm{C}_{1} \ldots \ldots . . .\left(2 \right ) &(\because \mathrm{p}=\mathrm{x}+2) \end{aligned}

         And, \int \frac{1}{\sqrt{x+2}} d x

Put  \mathrm{x}+2=\mathrm{U}=>\mathrm{d} \mathrm{x}=\mathrm{d} \mathrm{u}, then

\begin{aligned} \therefore \int \frac{1}{\sqrt{x+2}} d x &=\int \frac{1}{\sqrt{u}} d u=\int u^{-1 / 2} d u \\ &=\frac{u \frac{-1}{2}+1}{-1 / 2^{+1}}+C_{2} \quad \quad\left(\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right) \\ &=\frac{u \frac{-1}{2}+1}{-1 / 2^{+1}}+C_{2}=2 \sqrt{u}+C_{2} \\ &=2 \sqrt{x+2}+C_{2} \ldots \ldots . .(\mathrm{III}) \quad(\because u=x+2) \end{aligned}

 Also,  \int \frac{1}{(x-1) \sqrt{x+2}} d x

 Put x+2=t^{2}=>d x=2 t d t, then

\begin{aligned} \int \frac{1}{(x-1) \sqrt{x+2}} \mathrm{dx} &=\int \frac{1}{\left(t^{2}-2-1\right) \sqrt{t^{2}}} 2 \mathrm{t} \mathrm{d} \mathrm{t} \\ &=2 \int \frac{1}{\left(t^{2}-3\right) t} \mathrm{t} \mathrm{d} \mathrm{t} \\ &=2 \int \frac{1}{t^{2}-(\sqrt{3})^{2}} \mathrm{~d} \mathrm{t} \end{aligned}

    \begin{aligned} &=2 \cdot \frac{1}{2 . \sqrt{3}} \log \left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|+C_{3} \quad\left(\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right. \\ &=\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C_{3} \ldots .(\mathrm{IV}) \end{aligned}

Put the values of equ.(II), (III) and (IV) in (I), then,

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

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