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

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Question Number 169654 by CrispyXYZ last updated on 05/May/22
∫_0 ^1 ((3x^3 −x^2 +2x−4)/( (√(x^2 −3x+2)))) dx = ?
$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}}\:\mathrm{d}{x}\:=\:? \\ $$
Answered by floor(10²Eta[1]) last updated on 05/May/22
(√(x^2 −3x+2))=x+t ⇒−3x+2=2xt+t^2 ⇒x=((2−t^2 )/(2t+3))⇒dx=((−2t(2t+3)−2(2−t^2 ))/((2t+3)^2 ))dt =((−2t^2 −6t−4)/((2t+3)^2 ))dt x+t=((t^2 +3t+2)/(2t+3)) ∫_(√2) ^(−1) ((3(((2−t^2 )/(2t+3)))^3 −(((2−t^2 )/(2t+3)))^2 +2(((2−t^2 )/(2t+3)))−4)/((t^2 +3t+2)/(2t+3)))(((−2t^2 −6t−4)/((2t+3)^2 )))dt 2∫_(−1) ^(√2) (((3(2−t^2 )^3 −(2−t^2 )^2 (2t+3)+2(2−t^2 )(2t+3)^2 −4(2t+3)^3 )/((2t+3)^3 )))(((2t+3)/(t^2 +3t+2)))(((t^2 +3t+2)/((2t+3)^2 )))dt 2∫_(−1) ^(√2) ((−3t^6 −2t^5 +7t^4 −48t^3 −170t^2 −176t−60)/((2t+3)^4 ))dt
$$\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{2}}=\mathrm{x}+\mathrm{t} \\ $$$$\Rightarrow−\mathrm{3x}+\mathrm{2}=\mathrm{2xt}+\mathrm{t}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{2}−\mathrm{t}^{\mathrm{2}} }{\mathrm{2t}+\mathrm{3}}\Rightarrow\mathrm{dx}=\frac{−\mathrm{2t}\left(\mathrm{2t}+\mathrm{3}\right)−\mathrm{2}\left(\mathrm{2}−\mathrm{t}^{\mathrm{2}} \right)}{\left(\mathrm{2t}+\mathrm{3}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\frac{−\mathrm{2t}^{\mathrm{2}} −\mathrm{6t}−\mathrm{4}}{\left(\mathrm{2t}+\mathrm{3}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$\mathrm{x}+\mathrm{t}=\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{3t}+\mathrm{2}}{\mathrm{2t}+\mathrm{3}} \\ $$$$\int_{\sqrt{\mathrm{2}}} ^{−\mathrm{1}} \frac{\mathrm{3}\left(\frac{\mathrm{2}−\mathrm{t}^{\mathrm{2}} }{\mathrm{2t}+\mathrm{3}}\right)^{\mathrm{3}} −\left(\frac{\mathrm{2}−\mathrm{t}^{\mathrm{2}} }{\mathrm{2t}+\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{\mathrm{2}−\mathrm{t}^{\mathrm{2}} }{\mathrm{2t}+\mathrm{3}}\right)−\mathrm{4}}{\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{3t}+\mathrm{2}}{\mathrm{2t}+\mathrm{3}}}\left(\frac{−\mathrm{2t}^{\mathrm{2}} −\mathrm{6t}−\mathrm{4}}{\left(\mathrm{2t}+\mathrm{3}\right)^{\mathrm{2}} }\right)\mathrm{dt} \\ $$$$\mathrm{2}\int_{−\mathrm{1}} ^{\sqrt{\mathrm{2}}} \left(\frac{\mathrm{3}\left(\mathrm{2}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{3}} −\left(\mathrm{2}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{2t}+\mathrm{3}\right)+\mathrm{2}\left(\mathrm{2}−\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{2t}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2t}+\mathrm{3}\right)^{\mathrm{3}} }{\left(\mathrm{2t}+\mathrm{3}\right)^{\mathrm{3}} }\right)\left(\frac{\mathrm{2t}+\mathrm{3}}{\mathrm{t}^{\mathrm{2}} +\mathrm{3t}+\mathrm{2}}\right)\left(\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{3t}+\mathrm{2}}{\left(\mathrm{2t}+\mathrm{3}\right)^{\mathrm{2}} }\right)\mathrm{dt} \\ $$$$\mathrm{2}\int_{−\mathrm{1}} ^{\sqrt{\mathrm{2}}} \frac{−\mathrm{3t}^{\mathrm{6}} −\mathrm{2t}^{\mathrm{5}} +\mathrm{7t}^{\mathrm{4}} −\mathrm{48t}^{\mathrm{3}} −\mathrm{170t}^{\mathrm{2}} −\mathrm{176t}−\mathrm{60}}{\left(\mathrm{2t}+\mathrm{3}\right)^{\mathrm{4}} }\mathrm{dt} \\ $$
Answered by MJS_new last updated on 06/May/22
∫_0 ^1 ((3x^3 −x^2 +2x−4)/( (√(x^2 −3x+2))))dx= [t=2x−3 → dx=(dt/2)] =(1/8)∫_(−3) ^(−1) ((3t^3 +25t^2 +77t+55)/( (√(t^2 −1))))dt= [u=t+(√(t^2 −1)) → dt=((√(t^2 −1))/u)du] =(1/(64))∫_(−3+2(√2)) ^(−1) ((3u^6 +50u^5 +317u^4 +540u^3 +317u^2 +50u+3)/u^4 )du= =∫_(−3+2(√2)) ^(−1) (((3u^2 )/(64))+((25u)/(32))+((317)/(64))+((135)/(64u))+((317)/(64u^2 ))+((25)/(32u^3 ))+(3/(64u^4 )))du= =[(u^3 /(64))+((25u^2 )/(64))+((317u)/(64))+((135ln ∣u∣)/(16))−((317)/(64u))−((25)/(64u^2 ))−(1/(64u^3 ))]_(−3+2(√2)) ^(−1) = =[(((u−1)(u+1)(u^4 +25u^3 +318u^2 +25u+1))/(64u^3 ))+((135)/(16))ln ∣u∣]_(−3+2(√2)) ^(−1) = =−((101(√2)+135ln (−1+(√2)))/8)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{2}{x}−\mathrm{3}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\underset{−\mathrm{3}} {\overset{−\mathrm{1}} {\int}}\:\frac{\mathrm{3}{t}^{\mathrm{3}} +\mathrm{25}{t}^{\mathrm{2}} +\mathrm{77}{t}+\mathrm{55}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}={t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:\rightarrow\:{dt}=\frac{\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}{{u}}{du}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\underset{−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}} {\overset{−\mathrm{1}} {\int}}\:\frac{\mathrm{3}{u}^{\mathrm{6}} +\mathrm{50}{u}^{\mathrm{5}} +\mathrm{317}{u}^{\mathrm{4}} +\mathrm{540}{u}^{\mathrm{3}} +\mathrm{317}{u}^{\mathrm{2}} +\mathrm{50}{u}+\mathrm{3}}{{u}^{\mathrm{4}} }{du}= \\ $$$$=\underset{−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}} {\overset{−\mathrm{1}} {\int}}\:\left(\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{64}}+\frac{\mathrm{25}{u}}{\mathrm{32}}+\frac{\mathrm{317}}{\mathrm{64}}+\frac{\mathrm{135}}{\mathrm{64}{u}}+\frac{\mathrm{317}}{\mathrm{64}{u}^{\mathrm{2}} }+\frac{\mathrm{25}}{\mathrm{32}{u}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{64}{u}^{\mathrm{4}} }\right){du}= \\ $$$$=\left[\frac{{u}^{\mathrm{3}} }{\mathrm{64}}+\frac{\mathrm{25}{u}^{\mathrm{2}} }{\mathrm{64}}+\frac{\mathrm{317}{u}}{\mathrm{64}}+\frac{\mathrm{135ln}\:\mid{u}\mid}{\mathrm{16}}−\frac{\mathrm{317}}{\mathrm{64}{u}}−\frac{\mathrm{25}}{\mathrm{64}{u}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{64}{u}^{\mathrm{3}} }\right]_{−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}} ^{−\mathrm{1}} = \\ $$$$=\left[\frac{\left({u}−\mathrm{1}\right)\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{4}} +\mathrm{25}{u}^{\mathrm{3}} +\mathrm{318}{u}^{\mathrm{2}} +\mathrm{25}{u}+\mathrm{1}\right)}{\mathrm{64}{u}^{\mathrm{3}} }+\frac{\mathrm{135}}{\mathrm{16}}\mathrm{ln}\:\mid{u}\mid\right]_{−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}} ^{−\mathrm{1}} = \\ $$$$=−\frac{\mathrm{101}\sqrt{\mathrm{2}}+\mathrm{135ln}\:\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{8}} \\ $$

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