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

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Solution

$\int \frac{(x^{3} + 8) (x - 1)}{(x^{2} - 2 x + 4)} d x = \int \frac{(x^{3} + 2^{3}) (x - 1)}{(x^{2} - 2 x + 4)} d x = \int \frac{(x + 2) (x^{2} - 2 x + 4) (x - 1)}{(x^{2} - 2 x + 4)} d x [∴ a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})] = \int (x + 2) (x - 1) d x = \int (x^{2} - x + 2 x - 2) d x = \int (x^{2} + x - 2) d x = \int x^{2} d x + \int x d x - 2 \int 1 d x = \frac{x^{3}}{3} + \frac{x^{2}}{2} - 2 x + C$

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