dx-x-2-3x-4-

Question Number 125053 by bramlexs22 last updated on 08/Dec/20

$$\:\:\int\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}}\:=? \\ $$

Commented by Dwaipayan Shikari last updated on 08/Dec/20

∫(dx/( (√((x+(3/2))^2 −((5/2))^2 )))) x+(3/2)=(5/2)secθ⇒=(5/2)secθtanθ(dθ/dx) (5/2)∫((secθtanθdθ)/(tanθ)) =(5/2)log(secθ+tanθ)=(5/2)log(((2x+3)/5)+((√(4x^2 +12x−16))/5)) +C

$$\int\frac{{dx}}{\:\sqrt{\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} }}\:\:\:\:\:\:\:\:\:\:\:\:{x}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}}{sec}\theta\Rightarrow=\frac{\mathrm{5}}{\mathrm{2}}{sec}\theta{tan}\theta\frac{{d}\theta}{{dx}} \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}\int\frac{{sec}\theta{tan}\theta{d}\theta}{{tan}\theta}\:=\frac{\mathrm{5}}{\mathrm{2}}{log}\left({sec}\theta+{tan}\theta\right)=\frac{\mathrm{5}}{\mathrm{2}}{log}\left(\frac{\mathrm{2}{x}+\mathrm{3}}{\mathrm{5}}+\frac{\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{12}{x}−\mathrm{16}}}{\mathrm{5}}\right)\:+{C} \\ $$

Answered by benjo_mathlover last updated on 08/Dec/20

let (√(x^2 +3x−4)) = (x+4).t x−1 = (x+4) t^2 ⇒x = ((1+4t^2 )/(1−t^2 )) dx = ((10t)/((1−t^2 )^2 )) dt ∧ (√((x+4)(x−1))) = ((5t)/(1−t^2 )) I = ∫ (2/(1−t^2 )) dt = ℓn ∣((1+t)/(1−t)) ∣ + c I = ℓn ∣((1+(√((x−1)/(x+4))))/(1−(√((x−1)/(x+4))))) ∣ + c I = ℓn ∣(((√(x+4)) + (√(x−1)))/( (√(x+4)) −(√(x−1)))) ∣ + c

$${let}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}\:=\:\left({x}+\mathrm{4}\right).{t} \\ $$$$\:{x}−\mathrm{1}\:=\:\left({x}+\mathrm{4}\right)\:{t}^{\mathrm{2}} \:\Rightarrow{x}\:=\:\frac{\mathrm{1}+\mathrm{4}{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\:{dx}\:=\:\frac{\mathrm{10}{t}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:\wedge\:\sqrt{\left({x}+\mathrm{4}\right)\left({x}−\mathrm{1}\right)}\:=\:\frac{\mathrm{5}{t}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$${I}\:=\:\int\:\frac{\mathrm{2}}{\mathrm{1}−{t}^{\mathrm{2}} }\:{dt}\:=\:\ell{n}\:\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\:\mid\:+\:{c} \\ $$$${I}\:=\:\ell{n}\:\mid\frac{\mathrm{1}+\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{4}}}}{\mathrm{1}−\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{4}}}}\:\mid\:+\:{c}\: \\ $$$${I}\:=\:\ell{n}\:\mid\frac{\sqrt{{x}+\mathrm{4}}\:+\:\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{4}}\:−\sqrt{{x}−\mathrm{1}}}\:\mid\:+\:{c}\: \\ $$

Commented by bramlexs22 last updated on 08/Dec/20

$${Euler}\:{Substitution}\:{method} \\ $$

Answered by Bird last updated on 08/Dec/20

I=∫ (dx/( (√(x^2 +3x−4)))) x^2 +3x−4=0→Δ=9+16=25 ⇒x_1 =((−3+5)/2)=1 and x_2 =((−3−5)/2)=−4 ⇒I=∫ (dx/( (√((x−1)(x+4))))) we do the changement (√(x−1))=t ⇒x−1=t^2 ⇒I=∫ ((2tdt)/(t(√(t^2 +1+4)))) =2∫ (dt/( (√(t^2 +5)))) =_(t=(√5)u) 2∫ (((√5)du)/( (√5)(√(1+u^2 )))) =2∫ (du/( (√(1+u^2 ))))=2ln(u+(√(1+u^2 )))+c =2ln((t/( (√5)))+(√(1+(t^2 /5))))+c =2ln(((√(x−1))/( (√5)))+(√(1+((x−1)/5))))+C

$${I}=\int\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}}\: \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}=\mathrm{0}\rightarrow\Delta=\mathrm{9}+\mathrm{16}=\mathrm{25} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{3}+\mathrm{5}}{\mathrm{2}}=\mathrm{1}\:{and}\:{x}_{\mathrm{2}} =\frac{−\mathrm{3}−\mathrm{5}}{\mathrm{2}}=−\mathrm{4} \\ $$$$\Rightarrow{I}=\int\:\:\:\frac{{dx}}{\:\sqrt{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{4}\right)}} \\ $$$${we}\:{do}\:{the}\:{changement}\:\sqrt{{x}−\mathrm{1}}={t} \\ $$$$\Rightarrow{x}−\mathrm{1}={t}^{\mathrm{2}} \:\Rightarrow{I}=\int\:\:\frac{\mathrm{2}{tdt}}{{t}\sqrt{{t}^{\mathrm{2}} +\mathrm{1}+\mathrm{4}}} \\ $$$$=\mathrm{2}\int\:\:\frac{{dt}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{5}}}\:=_{{t}=\sqrt{\mathrm{5}}{u}} \:\:\mathrm{2}\int\:\frac{\sqrt{\mathrm{5}}{du}}{\:\sqrt{\mathrm{5}}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }} \\ $$$$=\mathrm{2}\int\:\:\frac{{du}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}=\mathrm{2}{ln}\left({u}+\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right)+{c} \\ $$$$=\mathrm{2}{ln}\left(\frac{{t}}{\:\sqrt{\mathrm{5}}}+\sqrt{\mathrm{1}+\frac{{t}^{\mathrm{2}} }{\mathrm{5}}}\right)+{c} \\ $$$$=\mathrm{2}{ln}\left(\frac{\sqrt{{x}−\mathrm{1}}}{\:\sqrt{\mathrm{5}}}+\sqrt{\mathrm{1}+\frac{{x}−\mathrm{1}}{\mathrm{5}}}\right)+{C} \\ $$$$ \\ $$

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