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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 21 maths textbook solution

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Answer:  \frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+c

Hint: Use substitution method to solve this integral.

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


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

        I=2 \int(2 x+1) \sqrt{t} \cdot \frac{d t}{(2 x+1)}=2 \int \sqrt{t} d t

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

        =2\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]+\mathrm{C}=2 \cdot \frac{2}{3} t^{\frac{3}{2}}+c

        =\frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+c \quad\left[\because t=x^{2}+x+1\right]

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