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Question Number 124532 by liberty last updated on 04/Dec/20

∫ (dx/(x^2 (√(x^2 −1)))) ?

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

Answered by bemath last updated on 04/Dec/20

ϑ(x) = ∫ (dx/(x^3 (√(1−x^(−2) )))) = ∫ (x^(−3) /( (√(1−x^(−2) )))) dx letting z = 1−x^(−2) →dz = 2x^(−3) dx ϑ(x)= (1/2)∫ (dz/( (√z))) = (1/2)∫ z^(−(1/2)) dz ϑ(x) = (√z) + c = (√(1−(1/x^2 ))) + c ϑ(x) = ((√(x^2 −1))/x) + c

$$\:\vartheta\left({x}\right)\:=\:\int\:\frac{{dx}}{{x}^{\mathrm{3}} \:\sqrt{\mathrm{1}−{x}^{−\mathrm{2}} }}\:=\:\int\:\frac{{x}^{−\mathrm{3}} }{\:\sqrt{\mathrm{1}−{x}^{−\mathrm{2}} }}\:{dx} \\ $$$${letting}\:{z}\:=\:\mathrm{1}−{x}^{−\mathrm{2}} \:\rightarrow{dz}\:=\:\mathrm{2}{x}^{−\mathrm{3}} \:{dx} \\ $$$$\:\vartheta\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{dz}}{\:\sqrt{{z}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:{z}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{dz}\: \\ $$$$\:\vartheta\left({x}\right)\:=\:\sqrt{{z}}\:+\:{c}\:=\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\:+\:{c}\: \\ $$$$\:\vartheta\left({x}\right)\:=\:\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}}\:+\:{c}\: \\ $$

Answered by MJS_new last updated on 04/Dec/20

I′m absolutely crazy and I should be kept away from students... this is no fun to transform but I had lots of fun transforming it and the integral... look at the integral... isn′t it super easy?! t=(√(2x(x+(√(x^2 −1))))) ⇔ x=(t^2 /(2(√(t^2 −1)))) → dx=((t(t^2 −2))/(2(t^2 −1)^(3/2) ))dt ∫(dx/(x^2 (√(x^2 −1))))=8∫(((t^2 −1)^(3/2) )/(t^4 (t^2 −2)))dx= = ♥ 4∫(dt/t^3 ) ♥ =−(2/t^2 )=−(1/(x(x+(√(x^2 −1)))))= =(((√(x^2 −1))−x)/x)=((√(x^2 −1))/x)−1=((√(x^2 −1))/x)+C

$$\mathrm{I}'\mathrm{m}\:\mathrm{absolutely}\:\mathrm{crazy}\:\mathrm{and}\:\mathrm{I}\:\mathrm{should}\:\mathrm{be}\:\mathrm{kept} \\ $$$$\mathrm{away}\:\mathrm{from}\:\mathrm{students}... \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{no}\:\mathrm{fun}\:\mathrm{to}\:\mathrm{transform}\:\mathrm{but}\:\mathrm{I}\:\mathrm{had}\:\mathrm{lots}\:\mathrm{of} \\ $$$$\mathrm{fun}\:\mathrm{transforming}\:\mathrm{it}\:\mathrm{and}\:\mathrm{the}\:\mathrm{integral}...\:\mathrm{look} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{integral}...\:\mathrm{isn}'\mathrm{t}\:\mathrm{it}\:{super}\:{easy}?! \\ $$$${t}=\sqrt{\mathrm{2}{x}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)}\:\Leftrightarrow\:{x}=\frac{{t}^{\mathrm{2}} }{\mathrm{2}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\rightarrow\:{dx}=\frac{{t}\left({t}^{\mathrm{2}} −\mathrm{2}\right)}{\mathrm{2}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }{dt} \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\mathrm{8}\int\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }{{t}^{\mathrm{4}} \left({t}^{\mathrm{2}} −\mathrm{2}\right)}{dx}= \\ $$$$=\:\heartsuit\:\mathrm{4}\int\frac{{dt}}{{t}^{\mathrm{3}} }\:\heartsuit\:=−\frac{\mathrm{2}}{{t}^{\mathrm{2}} }=−\frac{\mathrm{1}}{{x}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)}= \\ $$$$=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−{x}}{{x}}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}}−\mathrm{1}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}}+{C} \\ $$

Commented by bemath last updated on 04/Dec/20

hahaha..you are not crazy prof. but you are a master of math and santuy. wkwkwk...

$${hahaha}..{you}\:{are}\:{not}\:{crazy}\:{prof}.\:{but} \\ $$$${you}\:{are}\:{a}\:{master}\:{of}\:{math}\:{and}\: \\ $$$${santuy}.\:{wkwkwk}... \\ $$

Answered by Ar Brandon last updated on 04/Dec/20

I=∫(dx/(x^2 (√(x^2 −1))))=∫(dx/(x^3 (√(1−(1/x^2 ))))) 1−(1/x^2 )=u ⇒ (2/x^3 )dx=du I=∫(1/( (√u)))∙(du/2)=(√u)+C=(√(1−(1/x^2 )))+C

$$\mathcal{I}=\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}=\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} \sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{u}\:\Rightarrow\:\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{3}} }\mathrm{dx}=\mathrm{du} \\ $$$$\mathcal{I}=\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{u}}}\centerdot\frac{\mathrm{du}}{\mathrm{2}}=\sqrt{\mathrm{u}}+\mathcal{C}=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}+\mathcal{C} \\ $$

Answered by malwan last updated on 04/Dec/20

x = sec y ⇒ dx = secy tany dy ∴ ∫(( secy tany)/(sec^2 y (√(sec^2 y − 1)))) dy = ∫(( tan y)/(sec y (√(tan^2 y)))) dy = ∫(( dy)/(sec y)) = ∫ cos y dy = ((√(x^2 −1))/x) + C

$${x}\:=\:{sec}\:{y}\:\Rightarrow\:{dx}\:=\:{secy}\:{tany}\:{dy} \\ $$$$\therefore\:\int\frac{\:{secy}\:{tany}}{{sec}^{\mathrm{2}} {y}\:\sqrt{{sec}^{\mathrm{2}} {y}\:−\:\mathrm{1}}}\:{dy}\:=\:\int\frac{\:{tan}\:{y}}{{sec}\:{y}\:\sqrt{{tan}^{\mathrm{2}} {y}}}\:{dy} \\ $$$$=\:\int\frac{\:{dy}}{{sec}\:{y}}\:=\:\int\:{cos}\:{y}\:{dy}\:=\:\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}}\:+\:{C} \\ $$

Answered by mathmax by abdo last updated on 04/Dec/20

I =∫ (dx/(x^2 (√(x^2 −1)))) we do the changement x=cht ⇒ I =∫ ((sht)/(ch^2 t sh(t)))dt =2∫ (dt/(1+ch(2t))) =_(2t=u) 2∫ (du/(2(1+chu))) =∫ (du/(1+((e^(u ) +e^(−u) )/2))) =2 ∫ (du/(2+e^u +e^(−u) )) =_(e^u =y) 2∫ (dy/(y(2+y+y^(−1) ))) =2∫ (dy/(2y+y^2 +1)) =2∫ (dy/((y+1)^2 )) =−(2/(y+1)) +C we have y=e^u =e^(2t) and t=argch(x)=ln(x+(√(x^2 −1))) ⇒ e^(2t) =(x+(√(x^2 −1)))^2 ⇒ I =((−2)/((x+(√(x^2 −1)))^2 +1)) +C

$$\mathrm{I}\:=\int\:\:\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}=\mathrm{cht}\:\Rightarrow \\ $$$$\mathrm{I}\:=\int\:\:\:\frac{\mathrm{sht}}{\mathrm{ch}^{\mathrm{2}} \mathrm{t}\:\mathrm{sh}\left(\mathrm{t}\right)}\mathrm{dt}\:=\mathrm{2}\int\:\:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{ch}\left(\mathrm{2t}\right)}\:=_{\mathrm{2t}=\mathrm{u}} \:\:\mathrm{2}\int\:\:\:\frac{\mathrm{du}}{\mathrm{2}\left(\mathrm{1}+\mathrm{chu}\right)} \\ $$$$=\int\:\:\frac{\mathrm{du}}{\mathrm{1}+\frac{\mathrm{e}^{\mathrm{u}\:} +\mathrm{e}^{−\mathrm{u}} }{\mathrm{2}}}\:=\mathrm{2}\:\int\:\:\frac{\mathrm{du}}{\mathrm{2}+\mathrm{e}^{\mathrm{u}} \:+\mathrm{e}^{−\mathrm{u}} }\:=_{\mathrm{e}^{\mathrm{u}} \:=\mathrm{y}} \:\:\:\mathrm{2}\int\:\:\:\frac{\mathrm{dy}}{\mathrm{y}\left(\mathrm{2}+\mathrm{y}+\mathrm{y}^{−\mathrm{1}} \right)} \\ $$$$=\mathrm{2}\int\:\:\frac{\mathrm{dy}}{\mathrm{2y}+\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\:=\mathrm{2}\int\:\:\frac{\mathrm{dy}}{\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{2}}{\mathrm{y}+\mathrm{1}}\:+\mathrm{C}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{y}=\mathrm{e}^{\mathrm{u}} \:=\mathrm{e}^{\mathrm{2t}} \:\mathrm{and}\:\:\mathrm{t}=\mathrm{argch}\left(\mathrm{x}\right)=\mathrm{ln}\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow \\ $$$$\mathrm{e}^{\mathrm{2t}} \:=\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:\Rightarrow\:\mathrm{I}\:=\frac{−\mathrm{2}}{\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} +\mathrm{1}}\:+\mathrm{C} \\ $$

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