Mathematics

$$\displaystyle \int\frac{(x^{3} +8)(x-1)}{x^{2} -2x +4}dx$$


SOLUTION
$$\displaystyle I=\int \frac{(x^3+8)(x-1)}{(x^2-2x+4)}dx$$
$$\displaystyle =\int \frac{(x^3+(2)^3)(x-1)}{(x^2-2x+4)}dx$$
$$\displaystyle=\int\frac{(x+2)(x^2+4-2x)(x-1)}{(x^2-2x+4)}dx$$
$$\displaystyle= \int (x+2)(x-1)dx=\int x^2+x-2dx$$
$$=\dfrac{x^3}{3}+\dfrac{x^2}{2}-2x+c$$
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Subjective Medium Published on 17th 09, 2020
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