#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 13

Answer : -       $\frac{1}{2 \sqrt{33}} \log \left|\frac{\sqrt{11} x+\sqrt{3 x^{2}-12}}{\sqrt{11} x-\sqrt{3 x^{2}-12}}\right|+C$

Hint :-                Use substitution method and special integration formula.

Given :-                $\int \frac{1}{\left(2 x^{2}+3\right) \sqrt{x^{2}-4}} d x$

Sol : -

\begin{aligned} &\text { Let } \mathrm{I}=\int \frac{1}{\left(2 x^{2}+3\right) \sqrt{x^{2}-4}} d x \\ &\text { let, } \mathrm{x}=\frac{1}{t} \\ & \mathrm{dx}=-\frac{1}{t^{2}} \mathrm{dt} \text { then } \end{aligned}

$\begin{gathered} \quad I=\int \frac{1}{\left(2 \cdot \frac{1}{t^{2}}+3\right) \sqrt{\frac{1}{t^{2}}-4}}\left(-\frac{1}{t^{2}} \mathrm{dt}\right) \\ =-\int \frac{1}{\left(\frac{2+3 t^{2}}{t^{2}}\right) \sqrt{\frac{1-4 t^{2}}{t^{2}}}} \cdot \frac{1}{t^{2}} \mathrm{dt} \\ =-\int \frac{\frac{1}{t^{2}}}{\frac{1}{t^{2}}\left(3 t^{2}+2\right) \sqrt{1-4 t^{2}} \cdot \frac{1}{t}} \mathrm{dt} \\ \quad=-\int \frac{t}{\left(3 t^{2}+2\right) \sqrt{1-4 t^{2}}} \mathrm{dt} \end{gathered}$

\begin{aligned} &\text { Put } \text { . } 1-4 t^{2}=p^{2} \\ &\Rightarrow-4.2 \mathrm{t} \mathrm{dt}=2 \mathrm{p} \mathrm{dp} \\ &\Rightarrow \mathrm{tdt}=-\frac{p d p}{4} \end{aligned}

$\mathrm{I}=-\int \frac{1}{\left\{3\left(\frac{1-p^{2}}{4}\right)+2\right\} \cdot \sqrt{p^{2}}} \cdot-\frac{p d p}{4}\left[\because 1-4 t^{2}=p^{2} \rightarrow 1-p^{2}=4 t^{2} \rightarrow t^{2}=\frac{1-p^{2}}{4}\right]$

\begin{aligned} &\mathrm{I}=\frac{1}{4} \int \frac{1}{\frac{3-3 p^{2}+8}{4} \cdot p} \cdot \mathrm{pdp} \\ &\mathrm{I}=\frac{1}{4} \int \frac{4}{11-3 p^{2}} \mathrm{dp} \\ &\mathrm{I}=\frac{1}{3} \int \frac{1}{\left(\frac{11}{3}-p^{2}\right)} \mathrm{dp} \end{aligned}

\begin{aligned} &\mathrm{I}=\frac{1}{3} \int \frac{1}{\left(\sqrt{\frac{11}{3}}\right)^{2}-p^{2}} \mathrm{dp} \\ &\mathrm{I}=\frac{1}{3} \frac{1}{2 \cdot \frac{\sqrt{11}}{\sqrt{3}}} \log \left|\frac{\sqrt{\frac{11}{3}}+p}{\sqrt{\frac{11}{3}}-p}\right|+\mathrm{c}\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\right. \end{aligned}

\begin{aligned} &=\frac{1}{2 \sqrt{3} \cdot \sqrt{11}} \log \left|\frac{\sqrt{\frac{11}{3}}+\sqrt{1-4 t^{2}}}{\sqrt{\frac{11}{3}}-\sqrt{1-4 t^{2}}}\right|+c\left[\because p^{2}=1-4 t^{2} \rightarrow \mathrm{p}=\sqrt{1-4 t^{2}}\right]\\ &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\frac{\sqrt{11}+\sqrt{3} \sqrt{1-4 t^{2}}}{\sqrt{3}}}{\frac{\sqrt{11}-\sqrt{3} \sqrt{1-4 t^{2}}}{\sqrt{3}}}\right|+c \end{aligned}

\begin{aligned} &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\sqrt{11}+\sqrt{3\left(1-4 t^{2}\right)}}{\sqrt{11}-\sqrt{3\left(1-4 t^{2}\right)}}\right|+\mathrm{c} \\ &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\sqrt{11}+\sqrt{3\left(1-4 \frac{1}{x^{2}}\right)}}{\sqrt{11}-\sqrt{3\left(1-4 \frac{1}{x^{2}}\right)}}\right|+\mathrm{c}\left[\because \mathrm{x}=\frac{1}{t} \rightarrow \mathrm{t}=\frac{1}{\mathrm{~m}}\right] \end{aligned}

\begin{aligned} &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\sqrt{11}+\sqrt{\frac{3 x^{2}-12}{x^{2}}}}{\sqrt{11}-\sqrt{\frac{3 x^{2}-12}{x^{2}}}}\right|+\mathrm{c} \\ &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\sqrt{11}+\frac{1}{x} \sqrt{3 x^{2}-12}}{\sqrt{11}-\frac{1}{x} \sqrt{3 x^{2}-12}}\right|+\mathrm{c} \end{aligned}

\begin{aligned} &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\frac{\sqrt{11} x+\sqrt{3 x^{2}-12}}{x}}{\frac{\sqrt{\frac{11}{x}-\sqrt{3 x^{2}-12}}}{x}}\right|+\mathrm{c}\\ &=\frac{1}{2 \sqrt{33}} \log \left|\frac{\sqrt{11} x+\sqrt{3 x^{2}-12}}{\sqrt{11} x-\sqrt{3 x^{2}-12}}\right|+\mathrm{c} \end{aligned}