#### Please Solve RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.17 Question 6 Maths Textbook Solution

Answer:- $\frac{1}{\sqrt{2}}\sin^{-1}\left ( \frac{4x+3}{\sqrt{65}} \right )+c$

Hint:-to solve this problem, use special integration formula

Given:- $\int \frac{1}{\sqrt{7-3 x-2 x^{2}}} d x$

Solution:-

Let

\begin{aligned} &I=\int \frac{1}{\sqrt{7-3 x-2 x^{2}}} d x \Rightarrow \int \frac{1}{\sqrt{-2\left\{x^{2}+\frac{3}{2} x-7 / 2\right\}}}dx \\ &=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{x^{2}+2 x \cdot \frac{3}{4}+\left(\frac{3}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}-7 / 2\right\}}}dx \\ &=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{\left(x+\frac{3}{4}\right)^{2}-\frac{9}{16}-\frac{7}{2}\right\}}} d x \end{aligned}

\begin{aligned} &= \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{\left(x+\frac{3}{4}\right)^{2}-\frac{9+56}{16}\right\}}} d x \\ &= \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left\{\left(x+\frac{3}{4}\right)^{2}-\frac{65}{16}\right\}}} d x \Rightarrow \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{+\frac{65}{16}-\left(x+\frac{3}{4}\right)^{2}}} d x \\ &= \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(\frac{\sqrt{65}}{4}\right)^{2}-\left(x+\frac{3}{4}\right)^{2}}} d x \end{aligned}

\begin{aligned} &\text { put } \mathrm{x}+\frac{3}{4}=t \Rightarrow d x=d t \text { then } \\ &I=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left(\frac{\sqrt{65}}{4}\right)^{2}-t^{2}}} d t \Rightarrow \frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{t}{\left(\frac{\sqrt{65}}{4}\right)}\right)+c \quad\quad\quad\quad\left[\because \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right] \end{aligned}

\begin{aligned} &=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{x+3 / 4}{\frac{\sqrt{65}}{4}}\right)+c &\quad\quad\quad\quad{[\because t=x+3 / 4]}\\ &=\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{4 x+3}{\frac{4}{\sqrt{65}}}{4}\right) +c \Rightarrow \frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{4 x+3}{\sqrt{65}}\right)+c\\ \\ \end{aligned}