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  • please solve rd sharma class 12 chapter Indefinite Integrals exercise 18.2 question 4 maths textbook solution

please solve rd sharma class 12 chapter Indefinite Integrals exercise 18.2 question 4 maths textbook solution

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3 x^{4}+\frac{4}{3} x^{3}-\frac{17}{2} x^{2}+6 x+c

Hint: To solve this we multiply  (2-3 x)(3+2 x)(1-2 x)  then differentiate

\begin{aligned} &\text { Given: } \int(2-3 x)(3+2 x)(1-2 x) d x \\ &\text { Solution: }(2-3 x)(3+2 x)(1-2 x) \\ &=\left(6+4 x-9 x-6 x^{2}\right)(1-2 x) \\ &=\left(6-5 x-6 x^{2}\right)(1-2 x) \\ &=6-5 x-6 x^{2}+10 x^{2}-12 x+12 x^{3} \\ &=12 x^{3}+4 x^{2}-17 x+6 \end{aligned}

\begin{aligned} &\int(2-3 x)(3+2 x)(1-2 x) d x=\int\left(12 x^{3}+4 x^{2}-17 x+6\right) d x \\ &\text { Using identity } \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ &\frac{12 x^{4}}{3+1}+\frac{4 x^{3}}{3}-\frac{17 x^{2}}{1+1}+6 x+c \\ &=\frac{12 x^{4}}{4}+\frac{4 x^{3}}{3}-\frac{17 x^{2}}{2}+6 x+c \\ &=3 x^{4}+\frac{4}{3} x^{3}-\frac{17}{2} x^{2}+6 x+c \end{aligned}

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