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Question Number 211546 by MrGaster last updated on 12/Sep/24
∫(dx/( (√(x^2 −4x+13))))=?
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{\boldsymbol{{dx}}}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{x}}+\mathrm{13}}}=? \\ $$
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
$$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{13}}}\:\overset{\left[{t}=\frac{{x}−\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{13}}}{\mathrm{3}}\right]} {=}\: \\ $$$$=\int\frac{{dt}}{{t}}=\mathrm{ln}\:{t}\:=\mathrm{ln}\:\left({x}−\mathrm{2}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{13}}\right)\:+{C} \\ $$
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)
$$ \\ $$$$\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{euler}}\:\mathrm{1}^{\boldsymbol{\mathrm{st}}} \:\boldsymbol{\mathrm{sub}} \\ $$$$\boldsymbol{\mathrm{let}} \\ $$$$\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{x}}+\mathrm{13}}=\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{t}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{x}}+\mathrm{13}=\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{xt}}+\boldsymbol{\mathrm{t}}^{\mathrm{2}} \\ $$$$\left(−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}\right)\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{13} \\ $$$$\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{13}}{−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}},\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{t}}=\frac{\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{13}}{−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}}+\frac{\boldsymbol{\mathrm{t}}}{\mathrm{1}}=\frac{\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{13}−\mathrm{4}\boldsymbol{\mathrm{t}}−\mathrm{2}\boldsymbol{\mathrm{t}}^{\mathrm{2}} }{−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}} \\ $$$$=\frac{−\left(\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{t}}+\mathrm{13}\right)}{−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}} \\ $$$$\boldsymbol{\mathrm{dx}}=\frac{\left(−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}\right)\left(\mathrm{2}\boldsymbol{\mathrm{t}}\right)−\left(\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{13}\right)\left(−\mathrm{2}\right)}{\left(−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}\right)^{\mathrm{2}} }\boldsymbol{\mathrm{dt}} \\ $$$$\boldsymbol{\mathrm{dx}}=\frac{−\mathrm{8}\boldsymbol{\mathrm{t}}−\mathrm{4}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{26}}{\left(−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}\right)^{\mathrm{2}} }\boldsymbol{\mathrm{dt}}=\frac{−\mathrm{2}\boldsymbol{\mathrm{t}}^{\mathrm{2}} −\mathrm{8}\boldsymbol{\mathrm{t}}−\mathrm{26}}{\left(−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}\right)^{\mathrm{2}} }\boldsymbol{\mathrm{dt}} \\ $$$$=\frac{−\mathrm{2}\left(\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{t}}+\mathrm{13}\right)}{\left(−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}\right)^{\mathrm{2}} }\boldsymbol{\mathrm{dt}} \\ $$$$\int\frac{\frac{−\mathrm{2}\left(\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{t}}+\mathrm{13}\right)}{\left(−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}\right)^{\mathrm{2}} }\boldsymbol{\mathrm{dt}}}{\frac{−\left(\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{t}}+\mathrm{13}\right)}{−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}}}=\int\frac{−\mathrm{2}\left(\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{t}}+\mathrm{13}\right)\boldsymbol{\mathrm{dt}}}{\left(−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}\right)^{\mathrm{2}} }×\frac{−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}}{−\left(\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{t}}+\mathrm{13}\right)} \\ $$$$=\int\frac{\mathrm{2}}{−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}}×\boldsymbol{\mathrm{dt}} \\ $$$$\boldsymbol{\mathrm{u}}=−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}\:\:\:\frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{dt}}}=−\mathrm{2}\:\:\:\:\boldsymbol{\mathrm{dt}}=\frac{\boldsymbol{\mathrm{du}}}{−\mathrm{2}} \\ $$$$\int\frac{\mathrm{2}}{\boldsymbol{\mathrm{u}}}×\frac{\boldsymbol{\mathrm{du}}}{−\mathrm{2}}=−\boldsymbol{\mathrm{lnu}}=−\boldsymbol{\mathrm{ln}}\left(−\mathrm{4}−\mathrm{2}\boldsymbol{\mathrm{t}}\right)=\boldsymbol{\mathrm{ln}}\left(−\right)\left(\mathrm{4}+\mathrm{2}\boldsymbol{\mathrm{t}}\right) \\ $$$$=−\boldsymbol{\mathrm{ln}}\left(−\mathrm{4}−\mathrm{2}\left(\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{x}}+\mathrm{13}}+\boldsymbol{\mathrm{x}}\right)\right. \\ $$
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)))]
$$\mathrm{Error}\:\mathrm{in}\:\mathrm{line}\:\mathrm{3}\:\mathrm{of}\:\mathrm{the}\:\mathrm{calculation}: \\ $$$$\left(−\mathrm{4}−\mathrm{2}{t}\right){x}={t}^{\mathrm{2}} −\mathrm{13} \\ $$$$\left[…\mathrm{and}\:\mathrm{why}\:\mathrm{not}\:\mathrm{factorize}?\:\Rightarrow\:{x}=\frac{\mathrm{13}−{t}^{\mathrm{2}} }{\mathrm{2}\left({t}+\mathrm{2}\right)}\right] \\ $$
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
$$\boldsymbol{\mathrm{seen}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{man}},\boldsymbol{\mathrm{thanks}},\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{do}}\:\boldsymbol{\mathrm{love}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{break}} \\ $$$$\boldsymbol{\mathrm{things}}\:\boldsymbol{\mathrm{down}}\:\boldsymbol{\mathrm{also}}\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{others}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{comprehend}} \\ $$
Commented by Frix last updated on 14/Sep/24
Near the end: −ln (−4−2t) ≠ ln (4+2t)
$$\mathrm{Near}\:\mathrm{the}\:\mathrm{end}: \\ $$$$−\mathrm{ln}\:\left(−\mathrm{4}−\mathrm{2}{t}\right)\:\neq\:\mathrm{ln}\:\left(\mathrm{4}+\mathrm{2}{t}\right) \\ $$
Commented by klipto last updated on 14/Sep/24
thanks my man,frix told you then that let link up through whatsapp
$$\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{man}},\boldsymbol{\mathrm{frix}}\:\boldsymbol{\mathrm{told}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{that}} \\ $$$$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{link}}\:\boldsymbol{\mathrm{up}}\:\boldsymbol{\mathrm{through}}\:\boldsymbol{\mathrm{whatsapp}} \\ $$
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
$$\mathrm{It}\:\mathrm{would}\:\mathrm{be}\:\mathrm{easier}\:\mathrm{to}\:\mathrm{use} \\ $$$$\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{13}}={t}−{x}\:\Leftrightarrow\:{t}={x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{13}} \\ $$$$\Rightarrow \\ $$$${x}=\frac{{t}^{\mathrm{2}} −\mathrm{13}}{\mathrm{2}\left({t}−\mathrm{2}\right)} \\ $$$${dx}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{13}}}{{x}−\mathrm{2}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{13}}}{dt}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{13}}}{{t}−\mathrm{2}}{dt} \\ $$$$\Rightarrow \\ $$$$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{13}}}=\int\frac{{dt}}{{t}−\mathrm{2}}=\mathrm{ln}\:\left({t}−\mathrm{2}\right)\:= \\ $$$$=\mathrm{ln}\:\left({x}−\mathrm{2}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{13}}\right)\:+{C} \\ $$
Commented by klipto last updated on 14/Sep/24
cool also
$$\boldsymbol{\mathrm{cool}}\:\boldsymbol{\mathrm{also}} \\ $$

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