#### Need solution for RD Sharma maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.29 Question 11 maths textbook solution.

Answer : $\\I=\frac{1}{3}(x+3 x-18)^{3 / 2}-\frac{9}{8}(2 x+3) \sqrt{x^{2}+3 x-18}+\frac{729}{16} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}+3 x-18}\right|+C$

Hint: To solve the given integration,  we express the linear term as a derivative of quadratic into constant plus another constant

Given : $\int(x-3) \sqrt{x^{2}+3 x-18} d x$

Solution : $x-3=A+B \frac{d}{d x}\left(x^{2}+3 x-18\right)$

$\Rightarrow x-3=A+B(2 x+3)$

Comparing the coefficient of x and the constant terms, we get

$\Rightarrow 1=2 B \text { and } \Rightarrow-3=A+3 B$

$\Rightarrow B=1 / 2 \; \; \; \; \; \; \; \; \quad \Rightarrow-3-3 B=A$

\begin{aligned} &\Rightarrow-3-\frac{3}{2}=A \\ &\Rightarrow A=-\frac{9}{2} \end{aligned}

\begin{aligned} &I=\int\left[-\frac{9}{2}+\frac{1}{2}(2 x+3)\right] \sqrt{x^{2}+3 x-18} d x \\ &I=-\frac{9}{2} \int \sqrt{x^{2}+3 x-18} d x+\frac{1}{2} \int(2 x+3) \sqrt{x^{2}+3 x-18} d x \end{aligned}

For the second integral:

Let $x^{2}+3 x-18=t$

$\Rightarrow(2 x+3) d x=d t$

Use the formula : $\left[\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|\right]+C$

And $\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right]$

$\inline \\I=-\frac{9}{2}\left[\frac{\left(x+\frac{3}{2}\right)}{2} \sqrt{x^{2}+3 x-18}-\frac{81}{4 \times 2} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}+3 x-18}\right|\right]+\frac{1}{2} \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C$

\inline \begin{aligned} &I=-\frac{9}{2}\left[\left(\frac{2 x+3}{4}\right) \sqrt{x^{2}+3 x-18}-\frac{81}{8} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}+3 x-18}\right|+\frac{1}{3} t^{\frac{3}{2}}\right]+C \\ &I=\frac{1}{3}(x+3 x-18)^{3 / 2}-\frac{9}{8}(2 x+3) \sqrt{x^{2}+3 x-18}+\frac{729}{16} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}+3 x-18}\right|+C \end{aligned}