#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.21 Question 6 Maths Textbook Solution.

Answer: $-\sqrt{8+x-x^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{33}}\right)+c$

Given: $\int \frac{x}{\sqrt{8+x-x^{2}}} d x$

Hint: Simplify the given function

Solution:

\begin{aligned} &I=\int \frac{x}{\sqrt{8+x-x^{2}}} d x \\ &I=-\frac{1}{2} \int \frac{-2 x+1-1}{\sqrt{8+x-x^{2}}} d x \\ &I=-\frac{1}{2} \int \frac{-2 x+1}{\sqrt{8+x-x^{2}}} d x+\frac{1}{2} \int \frac{1}{\sqrt{8+x-x^{2}}} d x \end{aligned}

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

$I_{1}=-\frac{1}{2} \int \frac{-2 x+1}{\sqrt{8+x-x^{2}}} d x \text { and } I_{2}=\frac{1}{2} \int \frac{1}{\sqrt{8+x-x^{2}}} d x$

$I_{1}=-\frac{1}{2} \int \frac{-2 x+1}{\sqrt{8+x-x^{2}}} d x$

Let

$8+x-x^{2}=y$                                                        .............(2)

$(1-2 x) d x=d y$

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

\begin{aligned} &I_{1}=-\sqrt{y}+c \\ &I_{1}=-\sqrt{8+x-x^{2}}+c \end{aligned}                                    (From Equation 2)

Now,$I_{2}=\frac{1}{2} \int \frac{1}{\sqrt{8+x-x^{2}}} d x$

$I_{2}=\frac{1}{2} \int \frac{1}{\sqrt{8-\left(x^{2}-x+\frac{1}{4}\right)+\frac{1}{4}}} d x$

$I_{2}=\frac{1}{2} \int \frac{d x}{\sqrt{\frac{33}{4}-\left(x-\frac{1}{2}\right)^{2}}}$

$I_{2}=\frac{1}{2} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{33}}\right)+c$

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

$I=-\sqrt{8+x-x^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{33}}\right)+c$