Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

∫ (dx/( (√(x^2 +3x−4)))) =?

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

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

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

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com