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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∫−2x+3x2(x3+8)dx2)calculate∫1+∞−2x+3x2(x3+8)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....
letI=∫−2x+3x2(x3+8)dxchangementx=2tgiveI=∫−4t+34t2(8t3+8)2dt=−116∫4t−3t2(t3+1)dtletdecomposeF(t)=4t−3t2(t3+1)F(t)=4t−3t2(t+1)(t2−t+1)=at+bt2+ct+1+dt+et2−t+1b=limt→0t2F(t)=−3c=limt→−1(t+1)F(t)=−73⇒F(t)=at−3t2−73(t+1)+dt+et2−t+1limt→+∞tF(t)=0=a−73+d⇒d=−a+73F(t)=at−3t2−73(t+1)+(−a+73)t+et2−t+1F(1)=12=a−3−76−a+73+e⇒12=76+e⇒e=12−76=−46=−23⇒F(t)=at−3t2−73(t+1)+(−a+73)t−23t2−t+1F(2)=512(3)=a2−34−79+(−a+73)2−233=a2−27+2836−2a3+149−29=a2−5536−2a3+43=−a6+43−5536⇒5=−6a+36−55⇒6a=36−60=24⇒a=14F(t)=14t−3t2−73(t+1)+2512t−23t2−t+1⇒∫F(t)dt=14ln∣t∣+3t−73ln∣t+1∣+2512∫tdtt2−t+1−23∫dtt2−t+1but2512∫tdtt2−t+1=2524∫2t−1+3t2−t+1dt=2524ln∣t2−t+1∣+258∫dtt2−t+1and(−23+258)∫dtt2−t+1=38∫dtt2−t+1=38∫dt(t−12)2+34=t−12=32u3843∫11+u2du=12arctan(2t−13)⇒∫F(x)dx=14ln∣t∣+3t−73ln∣t+1∣+2524ln∣t2−t+1∣+12arctan(2t−13)+c….

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