#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 5 maths Textbook Solution.

Answer: $-3 \sqrt{5-x^{2}-2 x}-2 \sin ^{-1}\left(\frac{x+1}{\sqrt{6}}\right)+c$

Given: $\int \frac{3 x+1}{\sqrt{5-x^{2}-2 x}} d x$

Hint: Simplify it and then integrate

Solution:

\begin{aligned} &I=\int \frac{3 x+1}{\sqrt{5-x^{2}-2 x}} d x \\ &I=\int \frac{3 x+1}{\sqrt{6-\left(x^{2}+2 x+1\right)}} d x \\ &I=\int \frac{3 x+1}{\sqrt{6-(x+1)^{2}}} d x \\ &I=3 \int \frac{x+\frac{1}{3}}{\sqrt{6-(x+1)^{2}}} \end{aligned}

$=-\frac{3}{2} \int \frac{-2 x-\frac{2}{3}}{\sqrt{6-(x+1)^{2}}}$                                                    $\text { (multiplying and dividing numerator by-2 }$

\begin{aligned} &I=-\frac{3}{2} \int \frac{-2 x-2-\frac{2}{3}+2}{\sqrt{6-(x+1)^{2}}} d x \\ &I=-\frac{3}{2} \int \frac{-2 x-2}{\sqrt{5-x^{2}-2 x}}+-\frac{3}{2}\left(\frac{4}{3}\right) \int \frac{1}{\sqrt{6-(x+1)^{2}}} d x \end{aligned}

$I=I_{1}+I_{2}$             .................(1)

$I_{1}=-\frac{3}{2} \int \frac{-2 x-2}{\sqrt{5-x^{2}-2 x}} d x ; I_{2}=-2 \int \frac{1}{\sqrt{6-(x+1)^{2}}} d x$

Now,

Let

$5-x^{2}-2 x=y$                         ..........(2)

$(-2 x-2) d x=d y$

\begin{aligned} &\Rightarrow I_{1}=-\frac{3}{2} \int \frac{d x}{\sqrt{y}} \\ &I_{1}=-\frac{3}{2}\left(\frac{\sqrt{y}}{\frac{1}{2}}\right)+c \end{aligned}                    From Equation (2)

$I_{1}=-3 \sqrt{5-x^{2}-2 x}+c$

Now,$I_{2}=-2 \int \frac{1}{\sqrt{6-(x+1)^{2}}} d x$

$I_{2}=-2 \sin ^{-1}\left(\frac{x+1}{\sqrt{6}}\right)+c$

$\left[\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1} \frac{x}{a}+c\right]$

Putting value of $I_{1}$&$I_{2}$  in equation (1) and we get

$I=-3 \sqrt{5-x^{2}-2 x}-2 \sin ^{-1}\left(\frac{x+1}{\sqrt{6}}\right)+c$