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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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