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Question Number 60212 by peter frank last updated on 18/May/19

Commented by maxmathsup by imad last updated on 19/May/19

we have 3x^3 −x^2 +2x−4 =3x^3 −3x^2 +2x^2 +2x−4 =3x^2 (x−1) +2x^2 −2x +4x−4 =3x^2 (x−1) +2x(x−1)+4(x−1) =(x−1)(3x^2 +2x +4) also x^2 −3x +2 =x^2 −x −2x+2 =x(x−1)−2(x−1) =(x−1)(x−2) ⇒ I =∫_0 ^1 (((x−1)(3x^2 +2x+4))/((√(1−x))(√(2−x)))) dx =− ∫_0 ^1 (((√(1−x))(3x^2 +2x+4))/(√(2−x))) dx =−∫_0 ^1 (√((1−x)/(2−x)))(3x^2 +2x +4)dx changement (√((1−x)/(2−x)))=t give ((1−x)/(2−x)) =t^2 ⇒1−x =2t^2 −t^2 x ⇒1−2t^2 =(1−t^2 )x ⇒x =((1−2t^2 )/(1−t^2 )) ⇒ dx/dt =((−4t(1−t^2 )−(1−2t^2 )(−2t))/((1−t^2 )^2 )) =((−4t+4t^3 +2t−4t^3 )/((1−t^2 )^2 )) =((−2t)/((1−t^2 )^2 )) ⇒ I =−∫_(1/(√2)) ^0 t{ 3(((1−2t^2 )/(1−t^2 )))^2 +2((1−2t^2 )/(1−t^2 )) +4}(((−2t)/((1−t^2 )^2 )))dt I =∫_0 ^(1/(√2)) −2t^2 { ((3(1−2t^2 )^2 +2(1−2t^2 )(1−t^2 ) +4(1−t^2 )^2 )/((1−t^2 )^4 ))}dt I =−2 ∫_0 ^(1/(√2)) (t^2 /((1−t^2 )^4 )){ 3(4t^4 −4t^2 +1)+2(1−t^2 −2t^2 +2t^4 ) +4(t^4 −2t^2 +1)}dt =−2 ∫_0 ^(1/(√2)) (t^2 /((1−t^2 )^4 )){12t^4 −12t^2 +3 +2 −6t^2 +4t^4 +4t^4 −8t^2 +4}dt =−2 ∫_0 ^(1/(√2)) (t^2 /((1−t^2 )^4 )){ 20t^4 −26t^2 +9} dt =−2 ∫_0 ^(1/(√2)) ((20t^6 −26t^4 +9t^2 )/((1−t^2 )^4 )) dt let decompose F(t) =((20t^6 −26 t^4 +9t^2 )/((t^2 −1)^4 )) ⇒F(t) =(a/(t−1)) +(b/((t−1)^2 )) +(c/((t−1)^3 )) +(d/((t−1)^4 )) +(e/(t+1)) +(f/((t+1)^2 )) +(g/((t+1)^3 )) +(h/((t+1)^4 )) F(−t)=F(t) ⇒((−a)/(t+1)) +(b/((t+1)^2 )) −(c/((t+1)^3 )) +(d/((t+1)^4 )) −(e/(t−1)) +(f/((t−1)^2 )) −(g/((t−1)^3 )) +(h/((t−1)^4 )) =F(t) ⇒−a =e ,b=f , −c=g ,d =h ,....be continued...

$${we}\:{have}\:\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{4}\:=\mathrm{3}{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{4} \\ $$$$=\mathrm{3}{x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)\:+\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{4}{x}−\mathrm{4}\:=\mathrm{3}{x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)\:+\mathrm{2}{x}\left({x}−\mathrm{1}\right)+\mathrm{4}\left({x}−\mathrm{1}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}\:+\mathrm{4}\right)\:\:{also}\: \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}\:+\mathrm{2}\:={x}^{\mathrm{2}} −{x}\:−\mathrm{2}{x}+\mathrm{2}\:={x}\left({x}−\mathrm{1}\right)−\mathrm{2}\left({x}−\mathrm{1}\right)\:=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left({x}−\mathrm{1}\right)\left(\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{4}\right)}{\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{2}−{x}}}\:{dx}\:=−\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{\mathrm{1}−{x}}\left(\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{4}\right)}{\sqrt{\mathrm{2}−{x}}}\:{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}}\left(\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{4}\right){dx}\:\:\:{changement}\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}}={t}\:{give} \\ $$$$\frac{\mathrm{1}−{x}}{\mathrm{2}−{x}}\:={t}^{\mathrm{2}} \:\Rightarrow\mathrm{1}−{x}\:=\mathrm{2}{t}^{\mathrm{2}} −{t}^{\mathrm{2}} {x}\:\Rightarrow\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} =\left(\mathrm{1}−{t}^{\mathrm{2}} \right){x}\:\Rightarrow{x}\:=\frac{\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$${dx}/{dt}\:=\frac{−\mathrm{4}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)−\left(\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} \right)\left(−\mathrm{2}{t}\right)}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{−\mathrm{4}{t}+\mathrm{4}{t}^{\mathrm{3}} +\mathrm{2}{t}−\mathrm{4}{t}^{\mathrm{3}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{−\mathrm{2}{t}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${I}\:=−\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} ^{\mathrm{0}} \:{t}\left\{\:\mathrm{3}\left(\frac{\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\right)^{\mathrm{2}} \:+\mathrm{2}\frac{\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\:+\mathrm{4}\right\}\left(\frac{−\mathrm{2}{t}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right){dt} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} −\mathrm{2}{t}^{\mathrm{2}} \left\{\:\:\:\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\mathrm{2}\left(\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\:+\mathrm{4}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{4}} }\right\}{dt} \\ $$$${I}\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{4}} }\left\{\:\:\mathrm{3}\left(\mathrm{4}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{1}\right)+\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}{t}^{\mathrm{4}} \right)\:+\mathrm{4}\left({t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}\right)\right\}{dt} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \:\:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{4}} }\left\{\mathrm{12}{t}^{\mathrm{4}} \:−\mathrm{12}{t}^{\mathrm{2}} \:+\mathrm{3}\:+\mathrm{2}\:−\mathrm{6}{t}^{\mathrm{2}} \:+\mathrm{4}{t}^{\mathrm{4}} \:+\mathrm{4}{t}^{\mathrm{4}} \:−\mathrm{8}{t}^{\mathrm{2}} \:+\mathrm{4}\right\}{dt} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{4}} }\left\{\:\mathrm{20}{t}^{\mathrm{4}} \:−\mathrm{26}{t}^{\mathrm{2}} \:\:+\mathrm{9}\right\}\:{dt} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \:\frac{\mathrm{20}{t}^{\mathrm{6}} \:−\mathrm{26}{t}^{\mathrm{4}} \:+\mathrm{9}{t}^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{4}} }\:{dt}\:\:{let}\:{decompose}\: \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{20}{t}^{\mathrm{6}} −\mathrm{26}\:{t}^{\mathrm{4}} \:+\mathrm{9}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }\:\Rightarrow{F}\left({t}\right)\:=\frac{{a}}{{t}−\mathrm{1}}\:+\frac{{b}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} }\:+\frac{{d}}{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }\:+\frac{{e}}{{t}+\mathrm{1}} \\ $$$$+\frac{{f}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{g}}{\left({t}+\mathrm{1}\right)^{\mathrm{3}} }\:+\frac{{h}}{\left({t}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$${F}\left(−{t}\right)={F}\left({t}\right)\:\Rightarrow\frac{−{a}}{{t}+\mathrm{1}}\:+\frac{{b}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{{c}}{\left({t}+\mathrm{1}\right)^{\mathrm{3}} }\:+\frac{{d}}{\left({t}+\mathrm{1}\right)^{\mathrm{4}} }\:−\frac{{e}}{{t}−\mathrm{1}}\:+\frac{{f}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{{g}}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$+\frac{{h}}{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }\:={F}\left({t}\right)\:\Rightarrow−{a}\:={e}\:,{b}={f}\:,\:−{c}={g}\:\:,{d}\:={h}\:\:,....{be}\:{continued}... \\ $$

Answered by MJS last updated on 19/May/19

∫((3x^3 −x^2 +2x−4)/(√(x^2 −3x+2)))dx= [t=(√(x^2 −3x+2)) → dx=((2tdt)/(√(4t^2 +1)))] =∫(((25t^2 )/(√(4t^2 +1)))+((20)/(√(4t^2 +1)))+3t^2 +20)dt= =∫[(((25(8t^2 +1))/(8(√(4t^2 +1))))−((25)/(8(√(4t^2 +1)))))+((20)/(√(4t^2 +1)))+3t^2 +20]dt= =∫(((25(8t^2 +1))/(8(√(4t^2 +1))))+((135)/(8(√(4t^2 +1))))+3t^2 +20)dt ((25)/8)∫((8t^2 +1)/(√(4t^2 +1)))dt= [u=arcsinh 2t → dt=((√(4t^2 +1))/2)du] =((25)/(16))∫cosh 2u du=((25)/(32))sinh 2u =((25)/8)t(√(4t^2 +1))= =((25)/8)(2x−3)(√(x^2 −3x+2)) ((135)/8)∫(dt/(√(4t^2 +1)))= [u=arcsinh 2t → dt=((√(4t^2 +1))/2)du] =((135)/(16))∫du=((135)/(16))u=((135)/(16))arcsinh 2t = =((135)/(16))arcsinh 2(√(x^2 −3x+2)) ∫(3t^2 +20)dt=t^3 +20t=(x^3 −3x+22)(√(x^2 −3x+2)) ∫((3x^3 −x^2 +2x−4)/(√(x^2 −3x+2)))dx= =(1/8)(8x^2 +26x+101)(√(x^2 −3x+2))+((135)/(16))arcsinh 2(√(x^2 −3x+2)) +C

$$\int\frac{\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{tdt}}{\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}\right] \\ $$$$=\int\left(\frac{\mathrm{25}{t}^{\mathrm{2}} }{\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}+\frac{\mathrm{20}}{\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}+\mathrm{3}{t}^{\mathrm{2}} +\mathrm{20}\right){dt}= \\ $$$$=\int\left[\left(\frac{\mathrm{25}\left(\mathrm{8}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{8}\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}−\frac{\mathrm{25}}{\mathrm{8}\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}\right)+\frac{\mathrm{20}}{\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}+\mathrm{3}{t}^{\mathrm{2}} +\mathrm{20}\right]{dt}= \\ $$$$=\int\left(\frac{\mathrm{25}\left(\mathrm{8}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{8}\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}+\frac{\mathrm{135}}{\mathrm{8}\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}+\mathrm{3}{t}^{\mathrm{2}} +\mathrm{20}\right){dt} \\ $$$$ \\ $$$$\frac{\mathrm{25}}{\mathrm{8}}\int\frac{\mathrm{8}{t}^{\mathrm{2}} +\mathrm{1}}{\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{arcsinh}\:\mathrm{2}{t}\:\rightarrow\:{dt}=\frac{\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}}{du}\right] \\ $$$$=\frac{\mathrm{25}}{\mathrm{16}}\int\mathrm{cosh}\:\mathrm{2}{u}\:{du}=\frac{\mathrm{25}}{\mathrm{32}}\mathrm{sinh}\:\mathrm{2}{u}\:=\frac{\mathrm{25}}{\mathrm{8}}{t}\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{25}}{\mathrm{8}}\left(\mathrm{2}{x}−\mathrm{3}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}} \\ $$$$ \\ $$$$\frac{\mathrm{135}}{\mathrm{8}}\int\frac{{dt}}{\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{arcsinh}\:\mathrm{2}{t}\:\rightarrow\:{dt}=\frac{\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}}{du}\right] \\ $$$$=\frac{\mathrm{135}}{\mathrm{16}}\int{du}=\frac{\mathrm{135}}{\mathrm{16}}{u}=\frac{\mathrm{135}}{\mathrm{16}}\mathrm{arcsinh}\:\mathrm{2}{t}\:= \\ $$$$=\frac{\mathrm{135}}{\mathrm{16}}\mathrm{arcsinh}\:\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}} \\ $$$$ \\ $$$$\int\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{20}\right){dt}={t}^{\mathrm{3}} +\mathrm{20}{t}=\left({x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{22}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}} \\ $$$$ \\ $$$$\int\frac{\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{8}{x}^{\mathrm{2}} +\mathrm{26}{x}+\mathrm{101}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}+\frac{\mathrm{135}}{\mathrm{16}}\mathrm{arcsinh}\:\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}\:+{C} \\ $$

Commented by peter frank last updated on 19/May/19

thanks.sorry sir explain third line

$${thanks}.{sorry}\:{sir}\:{explain} \\ $$$${third}\:{line} \\ $$

Commented by MJS last updated on 19/May/19

I finished it

$$\mathrm{I}\:\mathrm{finished}\:\mathrm{it} \\ $$

Commented by maxmathsup by imad last updated on 19/May/19

thank you sir for this hardwork.

$${thank}\:{you}\:{sir}\:{for}\:{this}\:{hardwork}. \\ $$

Commented by peter frank last updated on 24/May/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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