Menu Close

dx-x-2-4x-13-

Tinku Tara 0 Comments
Pin




Question Number 211546 by MrGaster last updated on 12/Sep/24
∫(dx/( (√(x^2 −4x+13))))=?
∫dxx2−4x+13=?
Answered by Frix last updated on 12/Sep/24
∫(dx/( (√(x^2 −4x+13)))) =^([t=((x−2+(√(x^2 +4x+13)))/3)]) =∫(dt/t)=ln t =ln (x−2+(√(x^2 −4x+13))) +C
∫dxx2−4x+13=[t=x−2+x2+4x+133]=∫dtt=lnt=ln(x−2+x2−4x+13)+C
Answered by klipto last updated on 14/Sep/24
using euler 1^(st) sub let (√(x^2 −4x+13))=x+t x^2 −4x+13=x^2 +2xt+t^2 (−4−2t)x=t^2 −13 x=((t^2 −13)/(−4−2t)), x+t=((t^2 −13)/(−4−2t))+(t/1)=((t^2 −13−4t−2t^2 )/(−4−2t)) =((−(t^2 +4t+13))/(−4−2t)) dx=(((−4−2t)(2t)−(t^2 −13)(−2))/((−4−2t)^2 ))dt dx=((−8t−4t^2 +2t^2 −26)/((−4−2t)^2 ))dt=((−2t^2 −8t−26)/((−4−2t)^2 ))dt =((−2(t^2 +4t+13))/((−4−2t)^2 ))dt ∫((((−2(t^2 +4t+13))/((−4−2t)^2 ))dt)/((−(t^2 +4t+13))/(−4−2t)))=∫((−2(t^2 +4t+13)dt)/((−4−2t)^2 ))×((−4−2t)/(−(t^2 +4t+13))) =∫(2/(−4−2t))×dt u=−4−2t (du/dt)=−2 dt=(du/(−2)) ∫(2/u)×(du/(−2))=−lnu=−ln(−4−2t)=ln(−)(4+2t) =−ln(−4−2((√(x^2 −4x+13))+x)
usingeuler1stsubletx2−4x+13=x+tx2−4x+13=x2+2xt+t2(−4−2t)x=t2−13x=t2−13−4−2t,x+t=t2−13−4−2t+t1=t2−13−4t−2t2−4−2t=−(t2+4t+13)−4−2tdx=(−4−2t)(2t)−(t2−13)(−2)(−4−2t)2dtdx=−8t−4t2+2t2−26(−4−2t)2dt=−2t2−8t−26(−4−2t)2dt=−2(t2+4t+13)(−4−2t)2dt∫−2(t2+4t+13)(−4−2t)2dt−(t2+4t+13)−4−2t=∫−2(t2+4t+13)dt(−4−2t)2×−4−2t−(t2+4t+13)=∫2−4−2t×dtu=−4−2tdudt=−2dt=du−2∫2u×du−2=−lnu=−ln(−4−2t)=ln(−)(4+2t)=−ln(−4−2(x2−4x+13+x)
Commented by Frix last updated on 14/Sep/24
Error in line 3 of the calculation: (−4−2t)x=t^2 −13 [...and why not factorize? ⇒ x=((13−t^2 )/(2(t+2)))]
Errorinline3ofthecalculation:(−4−2t)x=t2−13[…andwhynotfactorize?⇒x=13−t22(t+2)]
Commented by klipto last updated on 14/Sep/24
seen my man,thanks,i do love to break things down also so that others can comprehend
seenmyman,thanks,idolovetobreakthingsdownalsosothatotherscancomprehend
Commented by Frix last updated on 14/Sep/24
Near the end: −ln (−4−2t) ≠ ln (4+2t)
Neartheend:−ln(−4−2t)≠ln(4+2t)
Commented by klipto last updated on 14/Sep/24
thanks my man,frix told you then that let link up through whatsapp
thanksmyman,frixtoldyouthenthatletlinkupthroughwhatsapp
Commented by Frix last updated on 14/Sep/24
It would be easier to use (√(x^2 −4x+13))=t−x ⇔ t=x+(√(x^2 −4x+13)) ⇒ x=((t^2 −13)/(2(t−2))) dx=((√(x^2 −4x+13))/(x−2+(√(x^2 −4x+13))))dt=((√(x^2 −4x+13))/(t−2))dt ⇒ ∫(dx/( (√(x^2 −4x+13))))=∫(dt/(t−2))=ln (t−2) = =ln (x−2+(√(x^2 −4x+13))) +C
Itwouldbeeasiertousex2−4x+13=t−x⇔t=x+x2−4x+13⇒x=t2−132(t−2)dx=x2−4x+13x−2+x2−4x+13dt=x2−4x+13t−2dt⇒∫dxx2−4x+13=∫dtt−2=ln(t−2)==ln(x−2+x2−4x+13)+C
Commented by klipto last updated on 14/Sep/24
cool also
coolalso

Pin

Leave a Reply

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