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

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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


            \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



                          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


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