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provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 22

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Answer: 2 \sqrt{2 x^{2}+3 x+1}+c

Hint:Use substitution method to solve this integral.

Given:   \int \frac{(4 x+3)}{\sqrt{2 x^{2}+3 x+1}} d x


        \begin{aligned} &\text { Let } I=\int \frac{(4 x+3)}{\sqrt{2 x^{2}+3 x+1}} d x \\ &\text { Put } 2 x^{2}+3 x+1=t \Rightarrow(4 x+3) d x=d t \text { then } \\ &I=\int \frac{1}{\sqrt{t}}=\int t^{\frac{-1}{2}} d t \end{aligned}

        =\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

        \begin{aligned} &=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c=2 \sqrt{t}+c \\ &=2 \sqrt{2 x^{2}+3 x+1}+c \quad \quad\left[\because t=2 x^{2}+3 x+1\right] \end{aligned}

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