dx-x-4x-2-2x-1-

Question Number 91842 by jagoll last updated on 03/May/20

$$\int\:\frac{{dx}}{{x}\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}\:? \\ $$

Commented by mathmax by abdo last updated on 03/May/20

I =∫ (dx/(x(√(4x^2 +2x−1)))) let solve 4x^2 +2x−1 =0 → Δ^′ =1+4 =5 ⇒x_1 =((−1+(√5))/4) and x_2 =((−1−(√5))/4) I =∫ (dx/(2x(√((x−x_1 )))(√(x−x_2 )))) we do the changement (√(x−x_1 ))=t ⇒ x−x_1 =t^2 ⇒I =∫ ((2tdt)/(2(t^2 +x_1 )t (√(t^2 +x_1 −x_2 )))) = ∫ (dt/((t^2 +x_1 )(√(t^2 +x_1 −x_2 )))) =∫ (dt/((t^2 +((−1+(√5))/2))(√(t^2 +((√5)/2))))) =_(t =(√((√5)/2))sh(t)) ∫ (1/((((√5)/2)sh^2 t +((−1+(√5))/2))(√((√5)/2))ch(t)))×(√((√5)/2))ch(t) dt =∫ ((2dt)/( (√5)sh^2 t +(√5)−1)) =∫ ((2dt)/( (√5)×((ch(2t)−1)/2)+(√5)−1)) =∫ ((4dt)/( (√5)ch(2t)−(√5)+2(√5)−2)) =∫ ((4dt)/( (√5)ch(2t)+(√5)−2)) =∫ ((4dt)/( (√5)×((e^(2t) +e^(−2t) )/2) +(√5)−2)) =∫ ((8dt)/( (√5)e^(2t) +(√5)e^(−2t) +2(√5)−4)) =_(e^(2t) =z) 8 ∫ (1/( (√5)z +(√5)z^(−1) +2(√5)−4))(dz/(2z)) =4 ∫ (dz/( (√5)z^2 +(2(√5)−4)z +(√5))) now its eazy to solve...

$${I}\:=\int\:\:\frac{{dx}}{{x}\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}\:\:{let}\:{solve}\:\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{1}\:=\mathrm{0}\:\rightarrow \\ $$$$\Delta^{'} \:=\mathrm{1}+\mathrm{4}\:=\mathrm{5}\:\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:{and}\:{x}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${I}\:=\int\:\:\frac{{dx}}{\mathrm{2}{x}\sqrt{\left({x}−{x}_{\mathrm{1}} \right)}\sqrt{{x}−{x}_{\mathrm{2}} }}\:\:{we}\:{do}\:{the}\:{changement}\:\sqrt{{x}−{x}_{\mathrm{1}} }={t}\:\Rightarrow \\ $$$${x}−{x}_{\mathrm{1}} ={t}^{\mathrm{2}} \:\Rightarrow{I}\:=\int\:\:\frac{\mathrm{2}{tdt}}{\mathrm{2}\left({t}^{\mathrm{2}} \:+{x}_{\mathrm{1}} \right){t}\:\sqrt{{t}^{\mathrm{2}} \:+{x}_{\mathrm{1}} −{x}_{\mathrm{2}} }} \\ $$$$=\:\int\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+{x}_{\mathrm{1}} \right)\sqrt{{t}^{\mathrm{2}} \:+{x}_{\mathrm{1}} −{x}_{\mathrm{2}} }}\:=\int\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\sqrt{{t}^{\mathrm{2}} +\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}} \\ $$$$=_{{t}\:=\sqrt{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}{sh}\left({t}\right)} \:\:\:\:\int\:\:\:\frac{\mathrm{1}}{\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}{sh}^{\mathrm{2}} {t}\:+\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\sqrt{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}{ch}\left({t}\right)}×\sqrt{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}{ch}\left({t}\right)\:{dt} \\ $$$$=\int\:\:\:\frac{\mathrm{2}{dt}}{\:\sqrt{\mathrm{5}}{sh}^{\mathrm{2}} {t}\:+\sqrt{\mathrm{5}}−\mathrm{1}}\:=\int\:\:\frac{\mathrm{2}{dt}}{\:\sqrt{\mathrm{5}}×\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{5}}−\mathrm{1}} \\ $$$$=\int\:\:\frac{\mathrm{4}{dt}}{\:\sqrt{\mathrm{5}}{ch}\left(\mathrm{2}{t}\right)−\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}}\:=\int\:\:\frac{\mathrm{4}{dt}}{\:\sqrt{\mathrm{5}}{ch}\left(\mathrm{2}{t}\right)+\sqrt{\mathrm{5}}−\mathrm{2}} \\ $$$$=\int\:\:\frac{\mathrm{4}{dt}}{\:\sqrt{\mathrm{5}}×\frac{{e}^{\mathrm{2}{t}} +{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\:+\sqrt{\mathrm{5}}−\mathrm{2}}\:=\int\:\frac{\mathrm{8}{dt}}{\:\sqrt{\mathrm{5}}{e}^{\mathrm{2}{t}} \:+\sqrt{\mathrm{5}}{e}^{−\mathrm{2}{t}} \:+\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4}} \\ $$$$=_{{e}^{\mathrm{2}{t}} \:={z}} \:\:\:\:\mathrm{8}\:\:\int\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}{z}\:+\sqrt{\mathrm{5}}{z}^{−\mathrm{1}} \:+\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4}}\frac{{dz}}{\mathrm{2}{z}} \\ $$$$=\mathrm{4}\:\int\:\:\frac{{dz}}{\:\sqrt{\mathrm{5}}{z}^{\mathrm{2}} \:+\left(\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4}\right){z}\:+\sqrt{\mathrm{5}}}\:\:{now}\:{its}\:{eazy}\:{to}\:{solve}… \\ $$$$ \\ $$

Answered by MJS last updated on 03/May/20

∫(dx/(x(√(a^2 x^2 +2x−1))))= [x=(1/(1+(√(a^2 +1))sin t)) ⇔ t=arcsin ((x−1)/(x(√(a^2 +1)))) → dx=x(√(a^2 x^2 +2x−1))] =∫dt=t=arcsin ((x−1)/(x(√(a^2 +1)))) +C

$$\int\frac{{dx}}{{x}\sqrt{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{x}=\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\mathrm{sin}\:{t}}\:\Leftrightarrow\:{t}=\mathrm{arcsin}\:\frac{{x}−\mathrm{1}}{{x}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\:\rightarrow\:{dx}={x}\sqrt{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}\right] \\ $$$$=\int{dt}={t}=\mathrm{arcsin}\:\frac{{x}−\mathrm{1}}{{x}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\:+{C} \\ $$

Commented by jagoll last updated on 03/May/20

$${mister},\:{you}\:{always}\:{have}\: \\ $$$${an}\:{idea}\:{to}\:{solve}\:{it} \\ $$

Commented by MJS last updated on 03/May/20

I'm an integral lover ��

Commented by jagoll last updated on 03/May/20

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