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#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 5

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

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

Given :-                    $\int \frac{x}{(x-3) \sqrt{x+1}} d x$

Sol : -     Let   $I=\int \frac{x}{(x-3) \sqrt{x+1}} d x$

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

Now,$\int \frac{1}{\sqrt{x+1}} d x$

Let   $x+1=p=>d x=d p$. Then,

\begin{aligned} \therefore \int \frac{1}{\sqrt{x+1}} d x &=\int \frac{1}{\sqrt{p}} d p=\int p^{-1 / 2} \mathrm{dp} \\ &=\frac{p^{-1 / 2+1}}{-1 / 2^{+1}}+\mathrm{C}_{1} \quad\left(\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right) \\ &=\frac{p^{1 / 2}}{1 / 2}+\mathrm{C}_{1}=2 \sqrt{x+1}+\mathrm{C}_{1} \ldots \ldots_{1}(\mathrm{II}) \quad(\because \mathrm{p}=\mathrm{x}+1) \end{aligned}

And, $\int \frac{1}{(x-3) \sqrt{x+1}} d x$

Put $\mathrm{x}+1=\mathrm{t}^{2}=>\mathrm{d} \mathrm{x}=2 \mathrm{t} \mathrm{d} \mathrm{t}$. Then,

\begin{aligned} \therefore \int \frac{1}{(x-3) \sqrt{x+1}} d x=& \int \frac{1}{\left(t^{2}-1-3\right) \sqrt{t^{2}}} 2 \mathrm{t} \mathrm{d} \mathrm{t} \quad\left(\because \mathrm{x}=\mathrm{t}^{2}-1\right) \\ &=2 \int \frac{1}{\left(t^{2}-4\right) t} \mathrm{t} \mathrm{d} \mathrm{t} \\ &=2 \int \frac{1}{\left(t^{2}\right)-(2)^{2}} \mathrm{~d} \mathrm{t} \end{aligned}

$\begin{array}{ll} =2 \cdot \frac{1}{2.2} \log \left|\frac{t-2}{t+2}\right|+\mathrm{C}_{2} & \left(\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+\mathrm{C}\right) \\\\ \left.=\frac{1}{2} \log \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+\mathrm{C}_{2} \ldots \ldots . \text { III }\right) & (\because \mathrm{t}=\sqrt{x+1}) \end{array}$

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

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