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Question Number 76356 by mathmax by abdo last updated on 26/Dec/19

find f(x) =∫ (dt/(√(t^2 −xt +1))) with x real.

$${find}\:{f}\left({x}\right)\:=\int\:\:\frac{{dt}}{\sqrt{{t}^{\mathrm{2}} −{xt}\:+\mathrm{1}}}\:\:{with}\:{x}\:{real}. \\ $$

Commented by mathmax by abdo last updated on 28/Dec/19

t^2 −xt +1 →Δ=x^2 −4 case 1 ∣x∣<2 ⇒Δ<0 ⇒t^2 −xt+1 =t^2 −2(x/2)t +(x^2 /4) +1−(x^2 /4) =(t−(x/2))^2 +((4−x^2 )/4) we do the changement t−(x/2)=((√(4−x^2 ))/2) u⇒ f(x) =∫ (1/((√((4−x^2 )/4))((√(1+u^2 )))))((√(4−x^2 ))/2)du =∫ (du/(√(1+u^2 ))) =ln(u+(√(1+u^2 ))) +c =ln(((2t−x)/(√(4−x^2 )))+(√(1+(((2t−x)/(√(4−x^2 ))))^2 ))) +c case 2 ∣x∣>2 ⇒Δ>0 ⇒t_1 =((x+(√(x^2 −4)))/2) and t_2 =((x−(√(x^2 −4)))/2) f(x)=∫ (dt/(√((t−t_1 )(t−t_2 )))) [we do the changement (√(t−t_1 ))=u ⇒ t=u^2 +t_1 ⇒f(x) =∫ ((2udu)/(u(√(u^2 +t_1 −t_2 )))) =2 ∫ (du/(√(u^2 +(√(x^2 −4))))) =_(u=(x^2 −4)^(1/4) z) 2 ∫ (((x^2 −4)^(1/4) dz)/(√((√(x^2 −4))(z^2 +1)))) =2 ∫ (dz/(√(z^2 +1))) =2ln(z+(√(1+z^2 ))) +C =2ln((x^2 −4)^(−(1/4)) u +(√(1+(x^2 −4)^(−(1/2)) u^2 ))) +C =2ln{(x^2 −4)(√(t−t_1 )) +(√(1+(x^2 −4)^(−(1/2)) (t−t_1 ))))} +C

$${t}^{\mathrm{2}} −{xt}\:+\mathrm{1}\:\rightarrow\Delta={x}^{\mathrm{2}} −\mathrm{4} \\ $$$${case}\:\mathrm{1}\:\:\mid{x}\mid<\mathrm{2}\:\Rightarrow\Delta<\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} −{xt}+\mathrm{1}\:={t}^{\mathrm{2}} −\mathrm{2}\frac{{x}}{\mathrm{2}}{t}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:+\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\left({t}−\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{4}−{x}^{\mathrm{2}} }{\mathrm{4}}\:\:{we}\:{do}\:{the}\:{changement}\:{t}−\frac{{x}}{\mathrm{2}}=\frac{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{\mathrm{2}}\:{u}\Rightarrow \\ $$$${f}\left({x}\right)\:=\int\:\:\frac{\mathrm{1}}{\sqrt{\frac{\mathrm{4}−{x}^{\mathrm{2}} }{\mathrm{4}}}\left(\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right)}\frac{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{\mathrm{2}}{du}\:=\int\:\:\frac{{du}}{\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }} \\ $$$$={ln}\left({u}+\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right)\:+{c}\:={ln}\left(\frac{\mathrm{2}{t}−{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{t}−{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\right)^{\mathrm{2}} }\right)\:\:+{c} \\ $$$${case}\:\mathrm{2}\:\mid{x}\mid>\mathrm{2}\:\Rightarrow\Delta>\mathrm{0}\:\Rightarrow{t}_{\mathrm{1}} =\frac{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\:{and}\:{t}_{\mathrm{2}} =\frac{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$${f}\left({x}\right)=\int\:\:\frac{{dt}}{\sqrt{\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)}}\:\left[{we}\:{do}\:{the}\:{changement}\:\sqrt{{t}−{t}_{\mathrm{1}} }={u}\:\Rightarrow\right. \\ $$$${t}={u}^{\mathrm{2}} \:+{t}_{\mathrm{1}} \:\Rightarrow{f}\left({x}\right)\:=\int\:\:\frac{\mathrm{2}{udu}}{{u}\sqrt{{u}^{\mathrm{2}} \:+{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }} \\ $$$$=\mathrm{2}\:\int\:\:\frac{{du}}{\sqrt{{u}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}}\:=_{{u}=\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} {z}} \:\mathrm{2}\:\int\:\:\frac{\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} {dz}}{\sqrt{\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}} \\ $$$$=\mathrm{2}\:\int\:\:\frac{{dz}}{\sqrt{{z}^{\mathrm{2}} \:+\mathrm{1}}}\:=\mathrm{2}{ln}\left({z}+\sqrt{\mathrm{1}+{z}^{\mathrm{2}} }\right)\:+{C} \\ $$$$=\mathrm{2}{ln}\left(\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} {u}\:+\sqrt{\mathrm{1}+\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {u}^{\mathrm{2}} }\right)\:+{C} \\ $$$$\left.=\mathrm{2}{ln}\left\{\left({x}^{\mathrm{2}} −\mathrm{4}\right)\sqrt{{t}−{t}_{\mathrm{1}} }\:+\sqrt{\mathrm{1}+\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left({t}−{t}_{\mathrm{1}} \right)}\right)\right\}\:+{C} \\ $$$$ \\ $$

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