Evaluate the following integrals: (x^3+8)(x-1) x^2-2x+4 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{(\text{x}^3+8)(\text{x}-1)}{\text{x}^2-2\text{x}+4}\text{dx}$

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Answer

$\int\frac{(\text{x}^3+8)(\text{x}-1)}{(\text{x}^2-2\text{x}+4)}\text{dx}$
$=\int\frac{(\text{x}^3+2^3)(\text{x}-1)}{(\text{x}^2-2\text{x}+4)}\text{dx}$
$=\int\frac{(\text{x}+2)(\text{x}^2-2\text{x}+4)(\text{x}-1)}{(\text{x}^2-2\text{x}+4)}\text{dx}$ $\big[\therefore\ \text{a}^3+\text{b}^3=(\text{a + b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=\int(\text{x}+2)(\text{x}-1)\text{dx}$
$=\int(\text{x}^2-\text{x}+2\text{x}-2)\text{dx}$
$=(\text{x}^2+\text{x}-2)\text{dx}$
$=\int\text{x}^2\text{dx}+\int\text{x dx}-2\int1\text{dx}$
$=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{2}-2\text{x}+\text{C}$

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