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Question Number 168745 by MikeH last updated on 17/Apr/22

∫(1/(x+(√(x−1)))) dx = ??

$$\int\frac{\mathrm{1}}{{x}+\sqrt{{x}−\mathrm{1}}}\:{dx}\:=\:?? \\ $$

Commented by safojontoshtemirov last updated on 17/Apr/22

∫(dx/(x+(√(x−1))))= (√(x−1))=t ; x=t^2 +1 ;dx=2tdt ∫((2tdt)/(t^2 +t+1))=∫((2t+1)/(t^2 +t+1))dt−∫(1/(t^2 +t+1))dt=I_1 −I_2 I_1 =∫((d(t^2 +t+1))/(t^2 +t+1))dt=ln(t^2 +t+1)+C_1 =ln(x−1+(√(x−1))+1)+C_1 I_2 =∫(1/(t^2 +t+1))dt=∫(1/((t+(1/2))^2 +(3/4)))dt=(4/3)∫(1/((((t+(1/2))/( ((√3)/2))))^2 +1))dt= (4/3)∫(1/((((2t+1)/( (√3))))^2 +1))dt=(4/3)∙((√3)/2)∫((d(((2t+1)/( (√3)))))/((((2t+1)/( (√3))))^2 +1))= (1/( (√3)))arctg(((2t+1)/( (√3))))+C_2 =(2/( (√3)))arctg((2(√(x−1))+1)/( (√3)))+C_2 I_1 −I_2 =ln(x−1+(√(x−1))+1)−(2/( (√3)))arctg((2(√(x−1))+1)/( (√3)))+C

$$\int\frac{{dx}}{{x}+\sqrt{{x}−\mathrm{1}}}=\:\:\sqrt{{x}−\mathrm{1}}={t}\:\:;\:{x}={t}^{\mathrm{2}} +\mathrm{1}\:\:;{dx}=\mathrm{2}{tdt} \\ $$$$\int\frac{\mathrm{2}{tdt}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}=\int\frac{\mathrm{2}{t}+\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{dt}−\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{dt}={I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\int\frac{{d}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{dt}={ln}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)+{C}_{\mathrm{1}} ={ln}\left({x}−\mathrm{1}+\sqrt{{x}−\mathrm{1}}+\mathrm{1}\right)+{C}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{dt}=\int\frac{\mathrm{1}}{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}{dt}=\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\mathrm{1}}{\left(\frac{{t}+\frac{\mathrm{1}}{\mathrm{2}}}{\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)^{\mathrm{2}} +\mathrm{1}}{dt}= \\ $$$$\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\mathrm{1}}{\left(\frac{\mathrm{2}{t}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}}{dt}=\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\int\frac{{d}\left(\frac{\mathrm{2}{t}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)}{\left(\frac{\mathrm{2}{t}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}}= \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{arctg}\left(\frac{\mathrm{2}{t}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C}_{\mathrm{2}} =\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{arctg}\frac{\mathrm{2}\sqrt{{x}−\mathrm{1}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+{C}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} −{I}_{\mathrm{2}} ={ln}\left({x}−\mathrm{1}+\sqrt{{x}−\mathrm{1}}+\mathrm{1}\right)−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{arctg}\frac{\mathrm{2}\sqrt{{x}−\mathrm{1}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+{C} \\ $$$$ \\ $$

Answered by blackmamba last updated on 17/Apr/22

= ∫ ((x−(√(x−1)))/(x^2 −x+1)) dx = ∫(((1/2)(2x−1)+(1/2))/(x^2 −x+1)) dx−∫ ((√(x−1))/(x^2 −x+1)) dx =(1/2)ln (x^2 −x+1)+(1/2)∫(dx/((x−(1/2))^2 +(((√3)/2))^2 ))−∫((√(x−1))/(x^2 −x+1)) dx

$$=\:\int\:\frac{{x}−\sqrt{{x}−\mathrm{1}}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:{dx}\: \\ $$$$=\:\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}−\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:{dx}−\int\:\frac{\sqrt{{x}−\mathrm{1}}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }−\int\frac{\sqrt{{x}−\mathrm{1}}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:{dx}\: \\ $$

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