∫_(0 ) ^1 ((3x^3 −x^2 +2x−4)/(√(x^2 −3x+2))) dx


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Question Number 61855 by aliesam last updated on 10/Jun/19

$\int_{0}^{1} \frac{3 x^{3} - x^{2} + 2 x - 4}{\sqrt{x^{2} - 3 x + 2}} d x$
	Answered by MJS last updated on 10/Jun/19

	$\frac{3 x^{3} - x^{2} + 2 x - 4}{\sqrt{x^{2} - 3 x + 2}} = - (3 x^{2} + 2 x + 4) \sqrt{\frac{x - 1}{x - 2}} for x < 2$ $- \int (3 x^{2} + 2 x + 4) \sqrt{\frac{x - 1}{x - 2}} d x =$ $[t = \sqrt{\frac{x - 1}{x - 2}} \to d x = - 2 \sqrt{{(x - 2)}^{3} (x - 1)}]$ $= \int \frac{2 t^{2} (20 t^{4} - 26 t^{2} + 9)}{{(t^{2} - 1)}^{4}} d t =$ $= \int (\frac{3}{8 {(t - 1)}^{4}} + \frac{7}{2 {(t - 1)}^{3}} + \frac{185}{16 {(t - 1)}^{2}} + \frac{135}{16 (t - 1)} + \frac{3}{8 {(t + 1)}^{4}} - \frac{7}{2 {(t + 1)}^{3}} + \frac{185}{16 {(t + 1)}^{2}} - \frac{135}{16 (t + 1)}) d t =$ $= - \frac{1}{8 {(t - 1)}^{3}} - \frac{7}{4 {(t - 1)}^{2}} - \frac{185}{16 (t - 1)} + \frac{135}{16} \ln (t - 1) - \frac{1}{8 {(t + 1)}^{3}} + \frac{7}{4 {(t + 1)}^{2}} - \frac{185}{16 (t + 1)} - \frac{135}{16} \ln (t + 1) =$ $= - \frac{t (185 t^{4} - 312 t^{2} + 135)}{8 {(t^{2} - 1)}^{3}} + \frac{135}{16} \ln \frac{t - 1}{t + 1} =$ $= - \frac{1}{8} (8 x^{2} + 26 x + 101) \sqrt{x^{2} - 3 x + 2} + \frac{135}{16} \ln (2 x - 3 - 2 \sqrt{x^{2} - 3 x + 2}) + C$ $\int_{0}^{1} \frac{3 x^{3} - x^{2} + 2 x - 4}{\sqrt{x^{2} - 3 x + 2}} d x = - \frac{101 \sqrt{2}}{8} - \frac{135}{16} \ln (3 - 2 \sqrt{2})$
	Commented by aliesam last updated on 10/Jun/19

	$t h a n k s s i r b r i l l i a n t s o l u t i o n$
	Commented by MJS last updated on 10/Jun/19

	${you}^{'} re welcome$
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