Menu Close

find-2x-3-x-2-x-3-8-dx-2-calculate-1-2x-3-x-2-x-3-8-dx-

Tinku Tara 0 Comments
Pin




Question Number 39017 by maxmathsup by imad last updated on 01/Jul/18
find ∫ ((−2x+3)/(x^2 ( x^3 +8)))dx 2) calculate ∫_1 ^(+∞) ((−2x+3)/(x^2 (x^3 +8)))dx
$${find}\:\:\:\int\:\:\frac{−\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} \left(\:{x}^{\mathrm{3}} \:+\mathrm{8}\right)}{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{−\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{3}} \:+\mathrm{8}\right)}{dx} \\ $$
Commented by math khazana by abdo last updated on 08/Jul/18
let I = ∫ ((−2x+3)/(x^2 (x^3 +8))) d x changement x=2t give I = ∫ ((−4t +3)/(4t^2 (8t^3 +8))) 2dt = −(1/(16)) ∫ ((4t−3)/(t^2 (t^3 +1)))dt let decompose F(t)= ((4t−3)/(t^2 (t^3 +1))) F(t) = ((4t−3)/(t^2 (t+1)(t^2 −t +1))) =(a/t) +(b/t^2 ) +(c/(t+1)) +((dt +e)/(t^2 −t+1)) b =lim_(t→0) t^2 F(t)= −3 c=lim_(t→−1) (t+1)F(t)= ((−7)/3) ⇒ F(t)= (a/t) −(3/t^2 ) −(7/(3(t+1))) +((dt +e)/(t^2 −t +1)) lim_(t→+∞) t F(t) =0 =a −(7/3) +d ⇒ d=−a +(7/3) F(t) = (a/t) −(3/t^2 ) −(7/(3(t+1))) +(((−a+(7/3))t +e)/(t^2 −t +1)) F(1) =(1/2) = a−3 −(7/6) −a +(7/3) +e ⇒ (1/2) = (7/6) +e ⇒e=(1/2) −(7/6) =−(4/6) =((−2)/3) ⇒ F(t)= (a/t) −(3/t^2 ) −(7/(3(t+1))) + (((−a +(7/3))t−(2/3))/(t^2 −t +1)) F(2)= (5/(12(3))) = (a/2) −(3/4) −(7/9) +(((−a +(7/3))2−(2/3))/3) = (a/2) −((27+28)/(36)) −((2a)/3) +((14)/9) −(2/9) =(a/2) −((55)/(36)) −((2a)/3) +(4/3) =−(a/6) +(4/3) −((55)/(36)) ⇒ 5 =−6a +36 −55 ⇒6a =36 −60 =24 ⇒a=(1/4) F(t)= (1/(4t)) −(3/t^2 ) −(7/(3(t+1))) +((((25)/(12))t −(2/3))/(t^2 −t +1)) ⇒ ∫ F(t)dt = (1/4)ln∣t∣ +(3/t) −(7/3)ln∣t+1∣ +((25)/(12)) ∫ ((tdt)/(t^2 −t+1)) −(2/3) ∫ (dt/(t^2 −t +1)) but ((25)/(12)) ∫ ((tdt)/(t^2 −t +1)) = ((25)/(24)) ∫ ((2t −1 +3)/(t^2 −t +1))dt =((25)/(24))ln∣t^2 −t +1∣ +((25)/8) ∫ (dt/(t^2 −t +1)) and (−(2/3) +((25)/8)) ∫ (dt/(t^2 −t +1)) =(3/8) ∫ (dt/(t^2 −t +1)) =(3/8) ∫ (dt/((t−(1/2))^2 +(3/4))) =_(t−(1/2)=((√3)/2)u) (3/8) (4/3)∫ (1/(1+u^2 ))du =(1/2) arctan(((2t−1)/( (√3)))) ⇒ ∫F(x)dx =(1/4)ln∣t∣ +(3/t) −(7/3)ln∣t+1∣+((25)/(24))ln∣t^2 −t+1∣ +(1/2)arctan(((2t−1)/( (√3))))+c....
$${let}\:{I}\:=\:\int\:\:\:\:\frac{−\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{3}} \:+\mathrm{8}\right)}\:{d}\:{x}\:\:\:{changement}\:{x}=\mathrm{2}{t}\:{give} \\ $$$${I}\:=\:\int\:\:\:\frac{−\mathrm{4}{t}\:+\mathrm{3}}{\mathrm{4}{t}^{\mathrm{2}} \left(\mathrm{8}{t}^{\mathrm{3}} \:+\mathrm{8}\right)}\:\mathrm{2}{dt} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{16}}\:\:\int\:\:\frac{\mathrm{4}{t}−\mathrm{3}}{{t}^{\mathrm{2}} \left({t}^{\mathrm{3}} \:+\mathrm{1}\right)}{dt}\:\:{let}\:{decompose} \\ $$$${F}\left({t}\right)=\:\frac{\mathrm{4}{t}−\mathrm{3}}{{t}^{\mathrm{2}} \left({t}^{\mathrm{3}} +\mathrm{1}\right)} \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{4}{t}−\mathrm{3}}{{t}^{\mathrm{2}} \left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}\right)}\:=\frac{{a}}{{t}}\:+\frac{{b}}{{t}^{\mathrm{2}} }\:+\frac{{c}}{{t}+\mathrm{1}}\:+\frac{{dt}\:+{e}}{{t}^{\mathrm{2}} \:−{t}+\mathrm{1}} \\ $$$${b}\:={lim}_{{t}\rightarrow\mathrm{0}} {t}^{\mathrm{2}} \:{F}\left({t}\right)=\:−\mathrm{3} \\ $$$${c}={lim}_{{t}\rightarrow−\mathrm{1}} \left({t}+\mathrm{1}\right){F}\left({t}\right)=\:\frac{−\mathrm{7}}{\mathrm{3}}\:\Rightarrow \\ $$$${F}\left({t}\right)=\:\frac{{a}}{{t}}\:−\frac{\mathrm{3}}{{t}^{\mathrm{2}} }\:−\frac{\mathrm{7}}{\mathrm{3}\left({t}+\mathrm{1}\right)}\:+\frac{{dt}\:+{e}}{{t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}} \\ $$$${lim}_{{t}\rightarrow+\infty} {t}\:{F}\left({t}\right)\:=\mathrm{0}\:={a}\:−\frac{\mathrm{7}}{\mathrm{3}}\:\:+{d}\:\Rightarrow\:{d}=−{a}\:+\frac{\mathrm{7}}{\mathrm{3}} \\ $$$${F}\left({t}\right)\:=\:\frac{{a}}{{t}}\:−\frac{\mathrm{3}}{{t}^{\mathrm{2}} }\:−\frac{\mathrm{7}}{\mathrm{3}\left({t}+\mathrm{1}\right)}\:\:+\frac{\left(−{a}+\frac{\mathrm{7}}{\mathrm{3}}\right){t}\:+{e}}{{t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:=\:{a}−\mathrm{3}\:−\frac{\mathrm{7}}{\mathrm{6}}\:\:−{a}\:\:+\frac{\mathrm{7}}{\mathrm{3}}\:+{e}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{7}}{\mathrm{6}}\:+{e}\:\Rightarrow{e}=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{7}}{\mathrm{6}}\:=−\frac{\mathrm{4}}{\mathrm{6}}\:=\frac{−\mathrm{2}}{\mathrm{3}}\:\Rightarrow \\ $$$${F}\left({t}\right)=\:\frac{{a}}{{t}}\:−\frac{\mathrm{3}}{{t}^{\mathrm{2}} }\:−\frac{\mathrm{7}}{\mathrm{3}\left({t}+\mathrm{1}\right)}\:+\:\frac{\left(−{a}\:+\frac{\mathrm{7}}{\mathrm{3}}\right){t}−\frac{\mathrm{2}}{\mathrm{3}}}{{t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{2}\right)=\:\frac{\mathrm{5}}{\mathrm{12}\left(\mathrm{3}\right)}\:=\:\frac{{a}}{\mathrm{2}}\:−\frac{\mathrm{3}}{\mathrm{4}}\:−\frac{\mathrm{7}}{\mathrm{9}}\:+\frac{\left(−{a}\:+\frac{\mathrm{7}}{\mathrm{3}}\right)\mathrm{2}−\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{3}} \\ $$$$=\:\frac{{a}}{\mathrm{2}}\:−\frac{\mathrm{27}+\mathrm{28}}{\mathrm{36}}\:−\frac{\mathrm{2}{a}}{\mathrm{3}}\:\:+\frac{\mathrm{14}}{\mathrm{9}}\:−\frac{\mathrm{2}}{\mathrm{9}} \\ $$$$=\frac{{a}}{\mathrm{2}}\:−\frac{\mathrm{55}}{\mathrm{36}}\:−\frac{\mathrm{2}{a}}{\mathrm{3}}\:\:+\frac{\mathrm{4}}{\mathrm{3}}\:=−\frac{{a}}{\mathrm{6}}\:\:+\frac{\mathrm{4}}{\mathrm{3}}\:−\frac{\mathrm{55}}{\mathrm{36}}\:\Rightarrow \\ $$$$\mathrm{5}\:=−\mathrm{6}{a}\:\:+\mathrm{36}\:−\mathrm{55}\:\Rightarrow\mathrm{6}{a}\:=\mathrm{36}\:−\mathrm{60}\:=\mathrm{24}\:\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${F}\left({t}\right)=\:\frac{\mathrm{1}}{\mathrm{4}{t}}\:−\frac{\mathrm{3}}{{t}^{\mathrm{2}} }\:−\frac{\mathrm{7}}{\mathrm{3}\left({t}+\mathrm{1}\right)}\:+\frac{\frac{\mathrm{25}}{\mathrm{12}}{t}\:−\frac{\mathrm{2}}{\mathrm{3}}}{{t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}}\:\Rightarrow \\ $$$$\int\:\:{F}\left({t}\right){dt}\:=\:\frac{\mathrm{1}}{\mathrm{4}}{ln}\mid{t}\mid\:+\frac{\mathrm{3}}{{t}}\:−\frac{\mathrm{7}}{\mathrm{3}}{ln}\mid{t}+\mathrm{1}\mid\:+\frac{\mathrm{25}}{\mathrm{12}}\:\int\:\:\:\frac{{tdt}}{{t}^{\mathrm{2}} \:−{t}+\mathrm{1}} \\ $$$$−\frac{\mathrm{2}}{\mathrm{3}}\:\int\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}}\:\:{but} \\ $$$$\frac{\mathrm{25}}{\mathrm{12}}\:\int\:\:\:\:\frac{{tdt}}{{t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}}\:=\:\frac{\mathrm{25}}{\mathrm{24}}\:\int\:\frac{\mathrm{2}{t}\:−\mathrm{1}\:+\mathrm{3}}{{t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}}{dt} \\ $$$$=\frac{\mathrm{25}}{\mathrm{24}}{ln}\mid{t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}\mid\:\:+\frac{\mathrm{25}}{\mathrm{8}}\:\int\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}}\:{and} \\ $$$$\left(−\frac{\mathrm{2}}{\mathrm{3}}\:+\frac{\mathrm{25}}{\mathrm{8}}\right)\:\int\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}}\:=\frac{\mathrm{3}}{\mathrm{8}}\:\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:−{t}\:+\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}\:\int\:\:\:\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{t}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u}} \:\:\:\frac{\mathrm{3}}{\mathrm{8}}\:\frac{\mathrm{4}}{\mathrm{3}}\int\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\Rightarrow \\ $$$$\int{F}\left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{4}}{ln}\mid{t}\mid\:+\frac{\mathrm{3}}{{t}}\:−\frac{\mathrm{7}}{\mathrm{3}}{ln}\mid{t}+\mathrm{1}\mid+\frac{\mathrm{25}}{\mathrm{24}}{ln}\mid{t}^{\mathrm{2}} \:−{t}+\mathrm{1}\mid \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{c}…. \\ $$$$ \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *