find ∫_0 ^1 ((x^3 −3)/(√(x^2 −x +2)))dx


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Question Number 73478 by abdomathmax last updated on 13/Nov/19

$f i n d \int_{0}^{1} \frac{x^{3} - 3}{\sqrt{x^{2} - x + 2}} d x$
	Answered by MJS last updated on 13/Nov/19

	$\int \frac{x^{3} - 3}{\sqrt{x^{3} - x + 2}} d x = \int \frac{x^{3} - 3}{\sqrt{{(x - \frac{1}{2})}^{2} + \frac{7}{4}}} d x =$ $[t = \frac{2 x - 1}{\sqrt{7}} \to d x = \frac{\sqrt{7}}{2} d t]$ $= \frac{1}{8} \int \frac{7 \sqrt{7} t^{3} + 21 t^{2} + 3 \sqrt{7} t - 23}{\sqrt{t^{2} + 1}} d t =$ $[t = \sinh \ln u = \frac{u^{2} - 1}{2 u} \Rightarrow u = t + \sqrt{t^{2} + 1} \to d t = \frac{\sqrt{t^{2} + 1}}{t + \sqrt{t^{2} + 1}} d u = \frac{u^{2} + 1}{2 u^{2}} d u]$ $= \frac{1}{64} \int \frac{7 \sqrt{7} u^{6} + 42 u^{5} - 9 \sqrt{7} u^{4} - 268 u^{3} + 9 \sqrt{7} u^{2} + 42 u - 7 \sqrt{7}}{u^{4}} d u$ $and {it}^{'} s easy to solve this$
	Commented by abdomathmax last updated on 17/Nov/19

	$t h a n k y o u s i r .$
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