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

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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


            \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


           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

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