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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)))
$${calculate}\:\:\int_{\mathrm{1}} ^{\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)} \\ $$
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...
$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}^{\mathrm{3}} }\:+\frac{{dx}+{e}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}} \\ $$$${c}=\frac{\mathrm{1}}{\mathrm{3}}\:\:{lim}_{{x}\rightarrow+\infty} \:{xF}\left({x}\right)=\mathrm{0}\:={a}+{d}\:\Rightarrow{d}=−{a}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }\:+\frac{−{ax}+{e}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}} \\ $$$${f}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }\:=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{−{ax}\:+{e}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)}−\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }\:=…\:\Rightarrow \\ $$$$\frac{\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{5}} \:+\mathrm{2}{x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{3}} }{\mathrm{3}{x}^{\mathrm{6}} \left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)}\:=…\:\Rightarrow \\ $$$$\frac{−{x}+\mathrm{2}}{\mathrm{3}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)}\:=\frac{{a}}{{x}}+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{−{ax}+{e}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}} \\ $$$${b}\:=\frac{\mathrm{2}}{\mathrm{9}}\:\Rightarrow{f}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }\:=\frac{{a}}{{x}}+\frac{\mathrm{2}}{\mathrm{9}{x}^{\mathrm{2}} }\:+\frac{−{ax}\:+{e}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{{a}}{{x}}+\frac{\mathrm{2}}{\mathrm{9}{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }\:+\frac{−{ax}+{e}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}} \\ $$$${f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{3}}\:={a}+\frac{\mathrm{2}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{−{a}+{e}}{\mathrm{2}}\:\Rightarrow \\ $$$$\frac{{a}}{\mathrm{2}}\:+\frac{\mathrm{2}}{\mathrm{9}}\:+\frac{{e}}{\mathrm{2}}\:=\mathrm{0}\Rightarrow\frac{{a}+{e}}{\mathrm{2}}\:=−\frac{\mathrm{2}}{\mathrm{9}}\:\Rightarrow{a}+{e}=−\frac{\mathrm{4}}{\mathrm{9}} \\ $$$${f}\left(−\mathrm{1}\right)=−\frac{\mathrm{1}}{\mathrm{6}}=−{a}+\frac{\mathrm{2}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{{a}+{e}}{\mathrm{6}}\:\Rightarrow \\ $$$$−\frac{\mathrm{5}}{\mathrm{6}}{a}\:+\frac{{e}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{9}}\:=−\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow−\mathrm{5}{a}+{e}−\frac{\mathrm{2}}{\mathrm{3}}\:=−\mathrm{1}\:\Rightarrow \\ $$$$−\mathrm{5}{a}\:+{e}=−\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{5}{a}−{e}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{5}{a}−\left(−{a}−\frac{\mathrm{4}}{\mathrm{9}}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{6}{a}+\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{6}{a}\:=\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{9}}\:=−\frac{\mathrm{1}}{\mathrm{9}}\:\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{54}} \\ $$$${e}=−\frac{\mathrm{4}}{\mathrm{9}}−{a}\:=\frac{\mathrm{1}}{\mathrm{54}}−\frac{\mathrm{4}}{\mathrm{9}}\:=\frac{−\mathrm{23}}{\mathrm{54}}\:\Rightarrow \\ $$$${f}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{54}{x}}+\frac{\mathrm{2}}{\mathrm{9}{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }\:+\frac{\frac{{x}}{\mathrm{54}}−\frac{\mathrm{23}}{\mathrm{54}}}{{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{3}}\:\Rightarrow \\ $$$$\int\:{f}\left({x}\right){dx}\:=−\frac{\mathrm{1}}{\mathrm{54}}{ln}\mid{x}\mid−\frac{\mathrm{2}}{\mathrm{9}{x}}\:+\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{−\mathrm{3}+\mathrm{1}}{x}^{−\mathrm{3}+\mathrm{1}} \\ $$$$+\frac{\mathrm{1}}{\mathrm{54}}\:\int\:\:\frac{{x}−\mathrm{23}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}}{dx}….{be}\:{continued}… \\ $$$$ \\ $$

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