#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 13 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 the given Function

Solution:

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

Multiplying and dividing the numerator by -2,

\begin{aligned} I &=\frac{-3}{2} \int \frac{-2 x-\frac{2}{3}}{\sqrt{5-x^{2}-2 x}} d x \\ I &=\frac{-3}{2} \int \frac{-2 x-2+\frac{4}{3}}{\sqrt{5-x^{2}-2 x}} d x \\ I &=\frac{-3}{2} \int \frac{-2 x-2}{\sqrt{5-x^{2}-2 x}} d x-2 \int \frac{1}{\sqrt{5-x^{2}-2 x}} d x \\ I &=I_{1}+I_{2} \\ 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{5-x^{2}-2 x}} d x \end{aligned}

Now, $I_{1}=\frac{-3}{2} \int \frac{-2 x-2}{\sqrt{5-x^{2}-2 x}} d x$

Let

\begin{aligned} &5-x^{2}-2 x=y \\ &(-2-2 x) d x=d y \end{aligned}

$\Rightarrow I_{1}=\frac{-3}{2} \int \frac{d y}{\sqrt{y}}=\frac{-3}{2}\left(\frac{\sqrt{y}}{\frac{1}{2}}\right)+c$

\begin{aligned} &I_{1}=-3 \sqrt{y}+c \\ &I_{1}=-3 \sqrt{5-x^{2}-2 x}+c \end{aligned}

Now, $I_{2}=-2 \int \frac{1}{\sqrt{5-x^{2}-2 x}} d x$

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

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

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