Question Number 222415 by Tawa11 last updated on 25/Jun/25
Answered by MrGaster last updated on 26/Jun/25
![t=(√((1−x)/(2−x))) x=((1−2t^2 )/(1−t^2 )) dx=((−2t)/((1−t^2 )^2 ))dt x=0⇒t=((√2)/2),x=1⇒t=0 (√((1−x)(2−x)))=(t/(1−t^2 )) 2x^2 +2x−4=2(((1−2t^2 )/(1−t^2 )))+2(((1−2t^2 )/(1−t^2 )))−4 =((8t^4 −6t^2 )/((1−t^2 )^2 )) ((2x^2 +2x−4)/( (√((1−x)(2−x)))))=((8t^4 −6t^2 )/((1−t^2 )^2 ))∙((1−t^2 )/t)=((8t^3 −6t)/(1−t^2 )) ∫_((√2)/2) ^0 ((8t^3 −6t)/(1−t^2 ))∙((−2t)/((1−t^2 )^2 ))dx=∫_0 ^((√2)/2) ((2(8t^4 −6t^2 ))/((1−t^2 )^3 )) =2∫_0 ^((√2)/2) ((8/(1−t^2 ))−((10)/((1−t^2 )^2 ))+(2/((1−t^2 )^3 )))dt t=tanh u,dt=tanh^2 udu,1−t^2 =sech^2 u t=0⇒u=0,t=((√2)/2)⇒tanh u=((√2)/2)⇒cosh u=(√2)⇒u=cosh^(−1) ((√2)) (1/(1−t^2 ))=cosh^2 u,(1/((1−t^2 )^2 ))=cosh^4 u,(1/((1−t^2 )^3 ))=cosh^6 u 2∫_0 ^(cosh^(−1) ((√2))e) (8 tanh−10cosh^4 u+2cosh^6 u)sech^2 udu =2∫_0 ^(cosh^(−1) ((√2))) (8−10 tanh+2cosh^4 u)du cosh^2 u=((1+cosh 2u)/2),cosh^4 u=(((1+cosh 2u)/2))^2 =(1/4)(1+2 cosh 2u+((1+cosh 4u)/2))=(3/8)+(1/2)cosh 2u+(1/8)cosh 4u 8−10∙((1+cosh 2u)/2)+2((3/8)−(1/2)cosh 2u+(1/8)cosh 2u)=((15)/4)−4 cosh 2u+(1/4)cosh 4u 2∫_0 ^(cosh^(−1) ((√2))) (((15)/4)−4 cosh 2u+(1/4)cosh 4u)du =2[((15)/4)u−2 sinh 2u+(1/(16))sinh 4u]_0 ^(cosh^(−1) ((√2))) u_0 =cosh^(−1) ((√2)),cosh v_0 =(√2),sinh v_0 =1 sinh 2v_0 =2 cosh^2 v_0 −1=2∙2−1=3 sinh 4v_0 =2sinh 2v_0 cosh 2v_0 =2∙2(√2)∙3=12(√2) [((15)/4)v_0 −2∙2(√2)+(1/(16))∙12(√2)]=((15)/4)u_0 −4(√2)+((3(√5))/4)=((15)/4)v_0 −((13(√2))/4) u_0 =cosh^(−1) ((√2))=ln((√2)+1) ((15)/2)ln(1+(√2))−((13(√2))/2)](https://www.tinkutara.com/question/Q222416.png)
$${t}=\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}} \\ $$$${x}=\frac{\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$${dx}=\frac{−\mathrm{2}{t}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$${x}=\mathrm{0}\Rightarrow{t}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}},{x}=\mathrm{1}\Rightarrow{t}=\mathrm{0} \\ $$$$\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{2}−{x}\right)}=\frac{{t}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}=\mathrm{2}\left(\frac{\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\right)+\mathrm{2}\left(\frac{\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\right)−\mathrm{4} \\ $$$$=\frac{\mathrm{8}{t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}}{\:\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{2}−{x}\right)}}=\frac{\mathrm{8}{t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\centerdot\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}}=\frac{\mathrm{8}{t}^{\mathrm{3}} −\mathrm{6}{t}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} ^{\mathrm{0}} \frac{\mathrm{8}{t}^{\mathrm{3}} −\mathrm{6}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\centerdot\frac{−\mathrm{2}{t}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \frac{\mathrm{2}\left(\mathrm{8}{t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \left(\frac{\mathrm{8}}{\mathrm{1}−{t}^{\mathrm{2}} }−\frac{\mathrm{10}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{2}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\right){dt} \\ $$$${t}=\mathrm{tanh}\:{u},{dt}=\mathrm{tanh}^{\mathrm{2}} {udu},\mathrm{1}−{t}^{\mathrm{2}} =\mathrm{sech}^{\mathrm{2}} {u} \\ $$$${t}=\mathrm{0}\Rightarrow{u}=\mathrm{0},{t}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow\mathrm{tanh}\:{u}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow\mathrm{cosh}\:{u}=\sqrt{\mathrm{2}}\Rightarrow{u}=\mathrm{cosh}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }=\mathrm{cosh}^{\mathrm{2}} {u},\frac{\mathrm{1}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }=\mathrm{cosh}^{\mathrm{4}} {u},\frac{\mathrm{1}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{3}} }=\mathrm{cosh}^{\mathrm{6}} {u} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{cosh}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)\mathrm{e}} \left(\mathrm{8}\:\mathrm{tanh}−\mathrm{10cosh}^{\mathrm{4}} {u}+\mathrm{2cosh}^{\mathrm{6}} {u}\right)\mathrm{sech}^{\mathrm{2}} {udu} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{cosh}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)} \left(\mathrm{8}−\mathrm{10}\:\mathrm{tanh}+\mathrm{2cosh}^{\mathrm{4}} {u}\right){du} \\ $$$$\mathrm{cosh}^{\mathrm{2}} {u}=\frac{\mathrm{1}+\mathrm{cosh}\:\mathrm{2}{u}}{\mathrm{2}},\mathrm{cosh}^{\mathrm{4}} {u}=\left(\frac{\mathrm{1}+\mathrm{cosh}\:\mathrm{2}{u}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{2}\:\mathrm{cosh}\:\mathrm{2}{u}+\frac{\mathrm{1}+\mathrm{cosh}\:\mathrm{4}{u}}{\mathrm{2}}\right)=\frac{\mathrm{3}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cosh}\:\mathrm{2}{u}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cosh}\:\mathrm{4}{u} \\ $$$$\mathrm{8}−\mathrm{10}\centerdot\frac{\mathrm{1}+\mathrm{cosh}\:\mathrm{2}{u}}{\mathrm{2}}+\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cosh}\:\mathrm{2}{u}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cosh}\:\mathrm{2}{u}\right)=\frac{\mathrm{15}}{\mathrm{4}}−\mathrm{4}\:\mathrm{cosh}\:\mathrm{2}{u}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cosh}\:\mathrm{4}{u} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{cosh}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)} \left(\frac{\mathrm{15}}{\mathrm{4}}−\mathrm{4}\:\mathrm{cosh}\:\mathrm{2}{u}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cosh}\:\mathrm{4}{u}\right){du} \\ $$$$=\mathrm{2}\left[\frac{\mathrm{15}}{\mathrm{4}}{u}−\mathrm{2}\:\mathrm{sinh}\:\mathrm{2}{u}+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sinh}\:\mathrm{4}{u}\right]_{\mathrm{0}} ^{\mathrm{cosh}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)} \\ $$$${u}_{\mathrm{0}} =\mathrm{cosh}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right),\mathrm{cosh}\:{v}_{\mathrm{0}} =\sqrt{\mathrm{2}},\mathrm{sinh}\:{v}_{\mathrm{0}} =\mathrm{1} \\ $$$$\mathrm{sinh}\:\mathrm{2}{v}_{\mathrm{0}} =\mathrm{2}\:\mathrm{cosh}^{\mathrm{2}} {v}_{\mathrm{0}} −\mathrm{1}=\mathrm{2}\centerdot\mathrm{2}−\mathrm{1}=\mathrm{3} \\ $$$$\mathrm{sinh}\:\mathrm{4}{v}_{\mathrm{0}} =\mathrm{2sinh}\:\mathrm{2}{v}_{\mathrm{0}} \mathrm{cosh}\:\mathrm{2}{v}_{\mathrm{0}} =\mathrm{2}\centerdot\mathrm{2}\sqrt{\mathrm{2}}\centerdot\mathrm{3}=\mathrm{12}\sqrt{\mathrm{2}} \\ $$$$\left[\frac{\mathrm{15}}{\mathrm{4}}{v}_{\mathrm{0}} −\mathrm{2}\centerdot\mathrm{2}\sqrt{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{16}}\centerdot\mathrm{12}\sqrt{\mathrm{2}}\right]=\frac{\mathrm{15}}{\mathrm{4}}{u}_{\mathrm{0}} −\mathrm{4}\sqrt{\mathrm{2}}+\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{4}}=\frac{\mathrm{15}}{\mathrm{4}}{v}_{\mathrm{0}} −\frac{\mathrm{13}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${u}_{\mathrm{0}} =\mathrm{cosh}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)=\mathrm{ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$$\frac{\mathrm{15}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)−\frac{\mathrm{13}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 26/Jun/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
Commented by Ghisom last updated on 27/Jun/25
$$\mathrm{something}\:\mathrm{went}\:\mathrm{wrong}\:\&\:\mathrm{it}'\mathrm{s}\:\mathrm{also}\:\mathrm{very}\:\mathrm{complicated}… \\ $$
Answered by Ghisom last updated on 27/Jun/25
![∫ ((3x^3 −x^2 +2x−4)/( (√(x^2 −3x+2))))dx= =∫ (((x−1)(3x^2 +2x+4))/( (√((1−x)(2−x)))))dx= =−∫ (((3x^2 +2x+4)(√(1−x)))/( (√(2−x))))dx= [t=((√(1−x))/( (√(2−x)))) → dx=−2(√((1−x)(2−x)^3 ))] =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 ))+((135)/8)∫(dt/(t^2 −1))= =−((t(185t^4 −312t^2 +135))/(8(t^2 −1)^3 ))+((135)/(16))ln ((t−1)/(t+1)) = =(((8x^2 +26x+101)(√((1−x)(2−x))))/8)+((135)/8)ln ∣(√(2−x))−(√(1−x))∣ +C ⇒ answer is ((135)/8)ln (1+(√2)) −((101(√2))/8)](https://www.tinkutara.com/question/Q222450.png)
$$\int\:\frac{\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}}{dx}= \\ $$$$=\int\:\frac{\left({x}−\mathrm{1}\right)\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)}{\:\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{2}−{x}\right)}}{dx}= \\ $$$$=−\int\:\frac{\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{2}−{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{2}−{x}}}\:\rightarrow\:{dx}=−\mathrm{2}\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{2}−{x}\right)^{\mathrm{3}} }\right] \\ $$$$=\mathrm{2}\int\:\frac{{t}^{\mathrm{2}} \left(\mathrm{20}{t}^{\mathrm{4}} −\mathrm{26}{t}^{\mathrm{2}} +\mathrm{9}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=−\frac{{t}\left(\mathrm{185}{t}^{\mathrm{4}} −\mathrm{312}{t}^{\mathrm{2}} +\mathrm{135}\right)}{\mathrm{8}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{135}}{\mathrm{8}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}= \\ $$$$=−\frac{{t}\left(\mathrm{185}{t}^{\mathrm{4}} −\mathrm{312}{t}^{\mathrm{2}} +\mathrm{135}\right)}{\mathrm{8}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{135}}{\mathrm{16}}\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\:= \\ $$$$=\frac{\left(\mathrm{8}{x}^{\mathrm{2}} +\mathrm{26}{x}+\mathrm{101}\right)\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{2}−{x}\right)}}{\mathrm{8}}+\frac{\mathrm{135}}{\mathrm{8}}\mathrm{ln}\:\mid\sqrt{\mathrm{2}−{x}}−\sqrt{\mathrm{1}−{x}}\mid\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{135}}{\mathrm{8}}\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:−\frac{\mathrm{101}\sqrt{\mathrm{2}}}{\mathrm{8}} \\ $$
Commented by Tawa11 last updated on 27/Jun/25
$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}. \\ $$
Answered by Ghisom last updated on 27/Jun/25
![∫_0 ^1 ((3x^3 −x^2 +2x−4)/( (√(x^2 −3x+2))))dx= =−∫_0 ^1 (((3x^2 +2x+4)(√(1−x)))/( (√(2−x))))dx= [t=(√(1−x))+(√(2−x)) → dx=−((2(√(1−x))(√(2−x)))/t)dt] =(1/(32))∫_(1+(√2)) ^1 (3t^5 −50t^3 +317t−540t^(−1) +317t^(−3) −50t^(−5) +3t^(−7) )dt= =[(t^6 /(64))−((25t^4 )/(64))+((317t^2 )/(64))−((135ln t)/8)−((317t^(−2) )/(64))+((25t^(−4) )/(64))−(t^(−6) /(64))]_(1+(√2)) ^1 = =((135)/8)ln (1+(√2)) −((101(√2))/8)](https://www.tinkutara.com/question/Q222452.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}}{dx}= \\ $$$$=−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{2}−{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}−{x}}\:\rightarrow\:{dx}=−\frac{\mathrm{2}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{2}−{x}}}{{t}}{dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\underset{\mathrm{1}+\sqrt{\mathrm{2}}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{3}{t}^{\mathrm{5}} −\mathrm{50}{t}^{\mathrm{3}} +\mathrm{317}{t}−\mathrm{540}{t}^{−\mathrm{1}} +\mathrm{317}{t}^{−\mathrm{3}} −\mathrm{50}{t}^{−\mathrm{5}} +\mathrm{3}{t}^{−\mathrm{7}} \right){dt}= \\ $$$$=\left[\frac{{t}^{\mathrm{6}} }{\mathrm{64}}−\frac{\mathrm{25}{t}^{\mathrm{4}} }{\mathrm{64}}+\frac{\mathrm{317}{t}^{\mathrm{2}} }{\mathrm{64}}−\frac{\mathrm{135ln}\:{t}}{\mathrm{8}}−\frac{\mathrm{317}{t}^{−\mathrm{2}} }{\mathrm{64}}+\frac{\mathrm{25}{t}^{−\mathrm{4}} }{\mathrm{64}}−\frac{{t}^{−\mathrm{6}} }{\mathrm{64}}\right]_{\mathrm{1}+\sqrt{\mathrm{2}}} ^{\mathrm{1}} = \\ $$$$=\frac{\mathrm{135}}{\mathrm{8}}\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:−\frac{\mathrm{101}\sqrt{\mathrm{2}}}{\mathrm{8}} \\ $$
Commented by Tawa11 last updated on 27/Jun/25
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}. \\ $$