Question Number 159960 by tounghoungko last updated on 23/Nov/21
$$\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}}\:{dx}\:=?\: \\ $$
Answered by MJS_new last updated on 23/Nov/21
![∫_0 ^1 ((3x^3 −x^2 +2x−4)/( (√(x^2 −3x+2))))dx= =∫_0 ^1 (((x−1)(3x^2 +2x+4))/( (√((x−1)(x−2)))))dx= [t=(√((x−1)/(x−2))) → dx=−2(√((x−1)(x−2)^3 ))dt] =2∫_((√2)/2) ^0 ((t^2 (20t^4 −26t^2 +9))/((t^2 −1)^4 ))dt= =−2∫_0 ^((√2)/2) ((t^2 (20t^4 −26t^2 +9))/((t^2 −1)^4 ))dt= [Ostrogradski′s Method] =[ ((t(185t^4 −312t^2 +135))/(8(t^2 −1)^3 ))]_0 ^((√2)/2) −((135)/8)∫_0 ^((√2)/2) (dt/(t^2 −1))= =[ ((t(185t^4 −312t^2 +135))/(8(t^2 −1)^3 ))−((135)/(16))ln ((t−1)/(t+1))]_0 ^((√2)/2) = =−((101(√2))/8)+((135)/8)ln (1+(√2))](https://www.tinkutara.com/question/Q159993.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}}{dx}= \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\left({x}−\mathrm{1}\right)\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)}{\:\sqrt{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\frac{{x}−\mathrm{1}}{{x}−\mathrm{2}}}\:\rightarrow\:{dx}=−\mathrm{2}\sqrt{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)^{\mathrm{3}} }{dt}\right] \\ $$$$=\mathrm{2}\underset{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} {\overset{\mathrm{0}} {\int}}\frac{{t}^{\mathrm{2}} \left(\mathrm{20}{t}^{\mathrm{4}} −\mathrm{26}{t}^{\mathrm{2}} +\mathrm{9}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }{dt}= \\ $$$$=−\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} {\int}}\frac{{t}^{\mathrm{2}} \left(\mathrm{20}{t}^{\mathrm{4}} −\mathrm{26}{t}^{\mathrm{2}} +\mathrm{9}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=\left[\:\frac{{t}\left(\mathrm{185}{t}^{\mathrm{4}} −\mathrm{312}{t}^{\mathrm{2}} +\mathrm{135}\right)}{\mathrm{8}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }\right]_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} −\frac{\mathrm{135}}{\mathrm{8}}\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} {\int}}\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}= \\ $$$$=\left[\:\frac{{t}\left(\mathrm{185}{t}^{\mathrm{4}} −\mathrm{312}{t}^{\mathrm{2}} +\mathrm{135}\right)}{\mathrm{8}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{135}}{\mathrm{16}}\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\right]_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} = \\ $$$$=−\frac{\mathrm{101}\sqrt{\mathrm{2}}}{\mathrm{8}}+\frac{\mathrm{135}}{\mathrm{8}}\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$