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Question Number 33589 by abdo imad last updated on 19/Apr/18
1) decompose F(x) = (1/((x^2 +4)(x−3)^2 )) 2) calculate ∫_4 ^(+∞) (dx/((x^2 +4)(x−3)^2 )) .
$$\left.\mathrm{1}\right)\:{decompose}\:{F}\left({x}\right)\:=\:\:\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)\left({x}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{4}} ^{+\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:. \\ $$
Commented by math khazana by abdo last updated on 22/Apr/18
F(x)= (a/(x−3)) +(b/((x−3)^2 )) +((cx+d)/(x^2 +4)) b =lim_(x→3) (x−3)^2 F(x)= (1/(13)) lim_(x→+∞) xF(x)=0 =a+c ⇒c=−a ⇒ F(x)= (a/(x−3)) +(1/(13(x−3)^2 )) +((−ax +d)/(x^2 +4)) F(0) = (1/(36)) = ((−a)/3) + (1/(117)) +(d/4) ⇒1=−12a +((36)/(117)) +9d ⇒ −12a +9d = 1−((36)/(117)) = ((81)/(117)) F(2)= (1/8) = −a +(1/(13)) +((−2a+d)/8) ⇒ 1= −8a +(8/(13)) −2a +d =−10a +d +(8/(13)) ⇒ −10a +d = 1−(8/(13)) = (5/(13)) we get tbe system −12a +9d = ((81)/(117)) and −10a +d =(5/(13)) ⇒ −90a +9d = ((45)/(13)) ⇒ 78a = ((81)/(117)) −((45)/(13)) ⇒ a= (1/(78))( ((81)/(117)) −((45)/(13))).....be continued....
$${F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{3}}\:+\frac{{b}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$${b}\:={lim}_{{x}\rightarrow\mathrm{3}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} {F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{13}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:={a}+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{13}\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:+\frac{−{ax}\:+{d}}{{x}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$${F}\left(\mathrm{0}\right)\:=\:\frac{\mathrm{1}}{\mathrm{36}}\:=\:\frac{−{a}}{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{117}}\:\:+\frac{{d}}{\mathrm{4}}\:\Rightarrow\mathrm{1}=−\mathrm{12}{a}\:+\frac{\mathrm{36}}{\mathrm{117}}\:+\mathrm{9}{d} \\ $$$$\Rightarrow\:−\mathrm{12}{a}\:+\mathrm{9}{d}\:=\:\mathrm{1}−\frac{\mathrm{36}}{\mathrm{117}}\:=\:\frac{\mathrm{81}}{\mathrm{117}} \\ $$$${F}\left(\mathrm{2}\right)=\:\frac{\mathrm{1}}{\mathrm{8}}\:=\:−{a}\:+\frac{\mathrm{1}}{\mathrm{13}}\:+\frac{−\mathrm{2}{a}+{d}}{\mathrm{8}}\:\Rightarrow \\ $$$$\mathrm{1}=\:−\mathrm{8}{a}\:+\frac{\mathrm{8}}{\mathrm{13}}\:−\mathrm{2}{a}\:+{d}\:=−\mathrm{10}{a}\:+{d}\:+\frac{\mathrm{8}}{\mathrm{13}}\:\Rightarrow \\ $$$$−\mathrm{10}{a}\:+{d}\:=\:\mathrm{1}−\frac{\mathrm{8}}{\mathrm{13}}\:=\:\frac{\mathrm{5}}{\mathrm{13}}\:{we}\:{get}\:{tbe}\:{system} \\ $$$$−\mathrm{12}{a}\:+\mathrm{9}{d}\:=\:\frac{\mathrm{81}}{\mathrm{117}}\:{and}\:\:−\mathrm{10}{a}\:+{d}\:=\frac{\mathrm{5}}{\mathrm{13}}\:\Rightarrow \\ $$$$−\mathrm{90}{a}\:+\mathrm{9}{d}\:=\:\frac{\mathrm{45}}{\mathrm{13}}\:\Rightarrow\:\mathrm{78}{a}\:=\:\frac{\mathrm{81}}{\mathrm{117}}\:−\frac{\mathrm{45}}{\mathrm{13}}\:\Rightarrow \\ $$$${a}=\:\frac{\mathrm{1}}{\mathrm{78}}\left(\:\frac{\mathrm{81}}{\mathrm{117}}\:−\frac{\mathrm{45}}{\mathrm{13}}\right)…..{be}\:{continued}…. \\ $$

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