How would you integrate (x-x^2)/(2 cube root x) with a lower limit of -1 and upper limit of -8?

How would you integrate (x-x^2)/(2 cube root x) with a lower limit of -1 and upper limit of -8?

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\[\int\limits_{-8}^{-1} \frac{ x-x ^{2} }{ 2\sqrt[3]{x} } dx\]

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where did the 3/10 and 3/18 come from?

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