0-1-3x-3-x-2-2x-4-x-2-3x-2-dx-

Question Number 159960 by tounghoungko last updated on 23/Nov/21

\int_{0}^{1} \frac{3 x^{3} - x^{2} + 2 x - 4}{\sqrt{x^{2} - 3 x + 2}} d x = ?

Answered by MJS_new last updated on 23/Nov/21

\underset{0}{\int^{1}} \frac{3 x^{3} - x^{2} + 2 x - 4}{\sqrt{x^{2} - 3 x + 2}} d x =

= \underset{0}{\int^{1}} \frac{(x - 1) (3 x^{2} + 2 x + 4)}{\sqrt{(x - 1) (x - 2)}} d x =

[t = \sqrt{\frac{x - 1}{x - 2}} \to d x = - 2 \sqrt{(x - 1) {(x - 2)}^{3}} d t]

= 2 \underset{\frac{\sqrt{2}}{2}}{\int^{0}} \frac{t^{2} (20 t^{4} - 26 t^{2} + 9)}{{(t^{2} - 1)}^{4}} d t =

= - 2 \underset{0}{\int^{\frac{\sqrt{2}}{2}}} \frac{t^{2} (20 t^{4} - 26 t^{2} + 9)}{{(t^{2} - 1)}^{4}} d t =

[{Ostrogradski}^{'} s Method]

= {[\frac{t (185 t^{4} - 312 t^{2} + 135)}{8 {(t^{2} - 1)}^{3}}]}_{0}^{\frac{\sqrt{2}}{2}} - \frac{135}{8} \underset{0}{\int^{\frac{\sqrt{2}}{2}}} \frac{d t}{t^{2} - 1} =

= {[\frac{t (185 t^{4} - 312 t^{2} + 135)}{8 {(t^{2} - 1)}^{3}} - \frac{135}{16} \ln \frac{t - 1}{t + 1}]}_{0}^{\frac{\sqrt{2}}{2}} =

= - \frac{101 \sqrt{2}}{8} + \frac{135}{8} \ln (1 + \sqrt{2})

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