Mathematics

Evaluate the following integral:
$$\int { \left( 2-3x \right) \left( 3+2x \right) \left( 1-2x \right)  } dx$$


SOLUTION
We have $$\left(2-3x\right)\left(3+2x\right)\left(1-2x\right)$$

$$=\left(2\left(3+2x\right)-3x\left(3+2x\right)\right)\left(1-2x\right)$$

$$=\left(6+4x-9x-6{x}^{2}\right)\left(1-2x\right)$$

$$=\left(6-5x-6{x}^{2}\right)\left(1-2x\right)$$

$$=6\left(1-2x\right)-5x\left(1-2x\right)-6{x}^{2}\left(1-2x\right)$$

$$=6-12x-5x+10{x}^{2}-6{x}^{2}+12{x}^{3}$$

$$=12{x}^{3}+4{x}^{2}-17x+6$$

Now $$\displaystyle\int{\left(2-3x\right)\left(3+2x\right)\left(1-2x\right)dx}$$

$$=\displaystyle\int{\left(12{x}^{3}+4{x}^{2}-17x+6\right)dx}$$

We know that $$\displaystyle\int{{x}^{n}dx}=\dfrac{{x}^{n+1}}{n+1}+c$$

$$=12\dfrac{{x}^{4}}{4}+4\dfrac{{x}^{3}}{3}-17\dfrac{{x}^{2}}{2}+6x+c$$ where $$c$$ is the constant of integration

$$=3{x}^{4}+\dfrac{4}{3}{x}^{3}-\dfrac{17}{2}{x}^{2}+6x+c$$
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Subjective Medium Published on 17th 09, 2020
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