#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 14

Answer : -       $\frac{1}{2 \sqrt{5}} \log \left|\frac{\sqrt{x^{2}+9}-\sqrt{5}}{\sqrt{x^{2}+9}+\sqrt{5}}\right|+C$

Hint :-                Use substitution method and special integration formula.

Given :-                  $\int \frac{x}{\left(x^{2}+4\right) \sqrt{x^{2}+9}} d x$

Sol : -

\begin{aligned} &\text { Let } \mathrm{I}=\int \frac{x}{\left(x^{2}+4\right) \sqrt{x^{2}+9}} d x \\ &\text { put, } x^{2}+9=u^{2} \rightarrow 2 \mathrm{x} \mathrm{dx}=2 \mathrm{u} \mathrm{du} \\ &\quad \rightarrow \mathrm{x} \mathrm{dx}=\mathrm{u} \text { du than } \\ &\qquad \mathrm{I}=\int \frac{u d u}{\left(u^{2}-9+4\right) \sqrt{u^{2}}} \end{aligned}

\begin{aligned} &\mathrm{I}=\int \frac{u d u}{\left(u^{2}-5\right) u} \quad\left[\because x^{2}+9=u^{2} \rightarrow x^{2}=u^{2}-9\right] \\ &\mathrm{I}=\int \frac{u}{u^{2}-5} \mathrm{~d} \mathrm{u} \\ &\mathrm{I}=\int \frac{u}{u^{2}-(\sqrt{5})^{2}} \mathrm{~d} \mathrm{u} \end{aligned}

\begin{aligned} &\mathrm{I}=\frac{1}{2 \cdot \sqrt{5}} \log \left|\frac{u-\sqrt{5}}{u+\sqrt{5}}\right|+\mathrm{c} \quad\left[\because \int \frac{1}{x^{2}-a^{2}} \mathrm{dx}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c\right] \\ &\mathrm{I}=\frac{1}{2 \sqrt{5}} \log \left(\frac{\sqrt{x^{2}+9}-\sqrt{5}}{\sqrt{x^{2}+9}+\sqrt{5}}\right) +\mathrm{C} \quad\left[\ddot x^{2}+9=u^{2} \rightarrow \mathrm{u}=\sqrt{x^{2}+9}\right] \end{aligned}