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calculate-1-dx-x-3-x-2-2x-3-

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Question Number 76360 by mathmax by abdo last updated on 26/Dec/19
calculate ∫_1 ^∞ (dx/(x^3 (x^2 −2x+3)))
calculate∫1∞dxx3(x2−2x+3)
Commented by abdomathmax last updated on 29/Dec/19
let decompose F(x)=(1/(x^3 (x^2 −2x+3))) F(x)=(a/x)+(b/x^2 ) +(c/x^3 ) +((dx+e)/(x^2 −2x+3)) c=(1/3) lim_(x→+∞) xF(x)=0 =a+d ⇒d=−a ⇒ f(x)=(a/x) +(b/x^2 ) +(1/(3x^3 )) +((−ax+e)/(x^2 −2x+3)) f(x)−(1/(3x^3 )) =(a/x) +(b/x^2 ) +((−ax +e)/(x^2 −2x+3)) ⇒ (1/(x^3 (x^2 −2x+3)))−(1/(3x^3 )) =... ⇒ ((3x^3 −x^5 +2x^4 −3x^3 )/(3x^6 (x^2 −2x+3))) =... ⇒ ((−x+2)/(3x^2 (x^2 −2x+3))) =(a/x)+(b/x^2 ) +((−ax+e)/(x^2 −2x+3)) b =(2/9) ⇒f(x)−(1/(3x^3 )) =(a/x)+(2/(9x^2 )) +((−ax +e)/(x^2 −2x+3)) ⇒ f(x)=(a/x)+(2/(9x^2 )) +(1/(3x^3 )) +((−ax+e)/(x^2 −2x+3)) f(1)=(1/3) =a+(2/9)+(1/3) +((−a+e)/2) ⇒ (a/2) +(2/9) +(e/2) =0⇒((a+e)/2) =−(2/9) ⇒a+e=−(4/9) f(−1)=−(1/6)=−a+(2/9)−(1/3) +((a+e)/6) ⇒ −(5/6)a +(e/6)−(1/9) =−(1/6) ⇒−5a+e−(2/3) =−1 ⇒ −5a +e=−(1/3) ⇒5a−e =(1/3) ⇒5a−(−a−(4/9))=(1/3) ⇒6a+(4/9)=(1/3) ⇒6a =(1/3)−(4/9) =−(1/9) ⇒a=−(1/(54)) e=−(4/9)−a =(1/(54))−(4/9) =((−23)/(54)) ⇒ f(x)=−(1/(54x))+(2/(9x^2 )) +(1/(3x^3 )) +(((x/(54))−((23)/(54)))/(x^2 −2x +3)) ⇒ ∫ f(x)dx =−(1/(54))ln∣x∣−(2/(9x)) +(1/3)×(1/(−3+1))x^(−3+1) +(1/(54)) ∫ ((x−23)/(x^2 −2x+3))dx....be continued...
letdecomposeF(x)=1x3(x2−2x+3)F(x)=ax+bx2+cx3+dx+ex2−2x+3c=13limx→+∞xF(x)=0=a+d⇒d=−a⇒f(x)=ax+bx2+13x3+−ax+ex2−2x+3f(x)−13x3=ax+bx2+−ax+ex2−2x+3⇒1x3(x2−2x+3)−13x3=…⇒3x3−x5+2x4−3x33x6(x2−2x+3)=…⇒−x+23x2(x2−2x+3)=ax+bx2+−ax+ex2−2x+3b=29⇒f(x)−13x3=ax+29x2+−ax+ex2−2x+3⇒f(x)=ax+29x2+13x3+−ax+ex2−2x+3f(1)=13=a+29+13+−a+e2⇒a2+29+e2=0⇒a+e2=−29⇒a+e=−49f(−1)=−16=−a+29−13+a+e6⇒−56a+e6−19=−16⇒−5a+e−23=−1⇒−5a+e=−13⇒5a−e=13⇒5a−(−a−49)=13⇒6a+49=13⇒6a=13−49=−19⇒a=−154e=−49−a=154−49=−2354⇒f(x)=−154x+29x2+13x3+x54−2354x2−2x+3⇒∫f(x)dx=−154ln∣x∣−29x+13×1−3+1x−3+1+154∫x−23x2−2x+3dx….becontinued…

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