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Question Number 126000 by bramlexs22 last updated on 16/Dec/20

∫ (dx/((x−1)(√(x^2 −2x)))) ?

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

Answered by Dwaipayan Shikari last updated on 16/Dec/20

∫(1/((x−1)(√((x−1)^2 −1))))dx = sec^(−1) (x−1)+C

$$\int\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}}{dx}\:=\:{sec}^{−\mathrm{1}} \left({x}−\mathrm{1}\right)+{C} \\ $$

Commented by bramlexs22 last updated on 16/Dec/20

yes..thanks

$${yes}..{thanks} \\ $$

Answered by benjo_mathlover last updated on 16/Dec/20

μ = ∫ (dx/((x−1)(√((x−1)^2 −1)))) μ = ∫ ((d(x−1))/((x−1)(√((x−1)^2 −1)))) μ = ∫ (dX/(X (√(X^2 −1)))) [ X = sec w ] μ = ∫ ((sec w tan w)/(sec w tan w)) dw = w + c μ = sec^(−1) (X)+c = sec^(−1) (x−1)+c

$$\:\mu\:=\:\int\:\frac{{dx}}{\left({x}−\mathrm{1}\right)\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\:\mu\:=\:\int\:\frac{{d}\left({x}−\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\:\mu\:=\:\int\:\frac{{dX}}{{X}\:\sqrt{{X}^{\mathrm{2}} −\mathrm{1}}}\:\:\left[\:{X}\:=\:\mathrm{sec}\:{w}\:\right]\: \\ $$$$\:\mu\:=\:\int\:\frac{\mathrm{sec}\:{w}\:\mathrm{tan}\:{w}}{\mathrm{sec}\:{w}\:\mathrm{tan}\:{w}}\:{dw}\:=\:{w}\:+\:{c}\: \\ $$$$\:\mu\:=\:\mathrm{sec}^{−\mathrm{1}} \left({X}\right)+{c}\:=\:\mathrm{sec}^{−\mathrm{1}} \left({x}−\mathrm{1}\right)+{c}\: \\ $$

Answered by mathmax by abdo last updated on 16/Dec/20

I=∫ (dx/((x−1)(√(x^2 −2x)))) ⇒I =∫ (dx/((x−1)(√((x−1)^2 −1)))) we do tbe changement x−1 =ch(t) ⇒I =∫ ((sh(t)dt)/((ch(t).sht))dt =∫ (2/(e^t +e^(−t) ))dt =_(e^t =u) 2 ∫ (1/(u+u^(−1) ))(du/u) =∫ ((2du)/(u^2 +1)) =ln(u^2 +1) +C =ln(e^(2t) +1) +C t=argch(x−1) =ln(x−1+(√((x−1)^2 −1))) ⇒ e^t =x−1+(√((x−1)^2 −1)) ⇒e^(2t) =(x−1+(√((x−1)^2 −1)))^2 ⇒ I =ln(1+(x−1+(√((x−1)^2 −1)))^2 ) +C

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

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