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Question Number 83960 by john santu last updated on 08/Mar/20

∫ ((x−1)/(√(x^2 −x))) dx ?

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

Commented by mathmax by abdo last updated on 08/Mar/20

I =∫ ((x−1)/(√(x^2 −x)))dx ⇒ I =(1/2)∫ ((2x−1−1)/(√(x^2 −x)))dx =(1/2)∫ ((2x−1)/(√(x^2 −x)))dx −(1/2)∫ (dx/(√(x^2 −x))) we have ∫ ((2x−1)/(2(√(x^2 −x))))dx =(√(x^2 −x)) +c_1 ∫ (dx/(√(x^2 −x))) =∫ (dx/(√(x(x−1)))) =∫ (dx/((√x)(√(x−1)))) =_((√x)=t) ∫ ((2tdt)/(t(√(t^2 −1)))) =2 ∫ (dt/(√(t^2 −1))) =2ln(t+(√(t^2 −1))) +c_2 =2ln((√x)+(√(x−1))) +c_2 ⇒ I =(√(x^2 −x)) −ln((√x)+(√(x−1))) +C

$${I}\:=\int\:\frac{{x}−\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} −{x}}}{dx}\:\Rightarrow\:{I}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}{x}−\mathrm{1}−\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} −{x}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} −{x}}}{dx}\:\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} −{x}}}\:\:{we}\:{have} \\ $$$$\int\:\:\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −{x}}}{dx}\:=\sqrt{{x}^{\mathrm{2}} −{x}}\:+{c}_{\mathrm{1}} \\ $$$$\int\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} −{x}}}\:=\int\:\frac{{dx}}{\sqrt{{x}\left({x}−\mathrm{1}\right)}}\:=\int\:\:\frac{{dx}}{\sqrt{{x}}\sqrt{{x}−\mathrm{1}}}\:=_{\sqrt{{x}}={t}} \:\:\int\:\:\frac{\mathrm{2}{tdt}}{{t}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\mathrm{2}\:\int\:\:\frac{{dt}}{\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}\:\:=\mathrm{2}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)\:+{c}_{\mathrm{2}} =\mathrm{2}{ln}\left(\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}\right)\:+{c}_{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\sqrt{{x}^{\mathrm{2}} −{x}}\:−{ln}\left(\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}\right)\:+{C} \\ $$

Answered by john santu last updated on 08/Mar/20

Answered by Kunal12588 last updated on 08/Mar/20

x−1=A(d/dx)(x^2 −x)+B ⇒x−1=2Ax−(A−B) A=(1/2), A−B=1 ⇒B=−(1/2) I=(1/2)∫(( (d/dx)(x^2 −x))/(√(x^2 −x)))−(1/2)∫(dx/(√((x−(1/2))^2 −((1/2))^2 ))) I=(√(x^2 −x))−(1/2)log∣x−(1/2)+(√(x^2 −x))∣+c I=(√(x^2 −x))−(1/2)log∣2x−1+2(√(x^2 −x))∣+c I=(√(x^2 −x))−(1/2)log∣((√x))^2 +2(√x)(√(x−1))+((√(x−1)))^2 ∣+c I=(√(x^2 −x))−log∣(√x)+(√(x−1))∣+c

$${x}−\mathrm{1}={A}\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} −{x}\right)+{B} \\ $$$$\Rightarrow{x}−\mathrm{1}=\mathrm{2}{Ax}−\left({A}−{B}\right) \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}},\:\:{A}−{B}=\mathrm{1}\:\Rightarrow{B}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\:\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} −{x}\right)}{\sqrt{{x}^{\mathrm{2}} −{x}}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$${I}=\sqrt{{x}^{\mathrm{2}} −{x}}−\frac{\mathrm{1}}{\mathrm{2}}{log}\mid{x}−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}^{\mathrm{2}} −{x}}\mid+{c} \\ $$$${I}=\sqrt{{x}^{\mathrm{2}} −{x}}−\frac{\mathrm{1}}{\mathrm{2}}{log}\mid\mathrm{2}{x}−\mathrm{1}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} −{x}}\mid+{c} \\ $$$${I}=\sqrt{{x}^{\mathrm{2}} −{x}}−\frac{\mathrm{1}}{\mathrm{2}}{log}\mid\left(\sqrt{{x}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{{x}}\sqrt{{x}−\mathrm{1}}+\left(\sqrt{{x}−\mathrm{1}}\right)^{\mathrm{2}} \mid+{c} \\ $$$${I}=\sqrt{{x}^{\mathrm{2}} −{x}}−{log}\mid\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}\mid+{c} \\ $$

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