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Question Number 41586 by MJS last updated on 09/Aug/18

f(x)=(√(−3+(√((x+1)/(x−1))))) ∫f(x)=? ∫f^(−1) (x)=?

$${f}\left({x}\right)=\sqrt{−\mathrm{3}+\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}} \\ $$$$\int{f}\left({x}\right)=? \\ $$$$\int{f}^{−\mathrm{1}} \left({x}\right)=? \\ $$

Commented by prof Abdo imad last updated on 10/Aug/18

f(x)=y ⇔x=f^(−1) (y) ⇒(√(−3+(√((x+1)/(x−1))))) =y ⇒ (√((x+1)/(x−1))) =y^2 +3 ⇒ ((x+1)/(x−1)) =(y^2 +3)^2 ⇒ x+1=x(y^2 +3)^2 −(y^2 +3)^2 ⇒ (1−(y^2 +3)^2 )x=−1−(y^2 +3)^2 ⇒ x=(((y^2 +3)^2 +1)/((y^2 +3)^2 −1)) =1 +(2/((y^2 +3)^2 −1)) ⇒ f^(−1) (x)= 1+(2/((x^2 +3)^2 −1)) ⇒ ∫ f^(−1) (x)dx= x +2 ∫ (dx/((x^2 +3)^2 −1)) +c but ∫ ((2dx)/((x^2 +3)^2 −1)) = ∫ ((2dx)/((x^2 +4)(x^2 +2))) = ∫ {(1/(x^2 +2)) −(1/(x^2 +4))}dx=∫ (dx/(x^2 +2)) −∫ (dx/(x^2 +4)) ∫ (dx/(x^2 +2)) =_(x=(√2)t) ∫ (((√2)dt)/(2(1+t^2 ))) =(1/(√2)) arctan((x/(√2))) ∫ (dx/(x^2 +4)) =_(x=2t) ∫ ((2dt)/(4(1+t^2 ))) =(1/2) arctan((t/2))⇒ ∫ f^(−1) (x)dx = x +(1/(√2)) arctan((x/(√2)))+(1/2)arctan((x/2))+c

$${f}\left({x}\right)={y}\:\Leftrightarrow{x}={f}^{−\mathrm{1}} \left({y}\right)\:\Rightarrow\sqrt{−\mathrm{3}+\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}}\:={y}\:\Rightarrow \\ $$$$\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}\:\:={y}^{\mathrm{2}} \:+\mathrm{3}\:\Rightarrow\:\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\:=\left({y}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${x}+\mathrm{1}={x}\left({y}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} −\left({y}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\left(\mathrm{1}−\left({y}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} \right){x}=−\mathrm{1}−\left({y}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${x}=\frac{\left({y}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} \:+\mathrm{1}}{\left({y}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1}}\:=\mathrm{1}\:+\frac{\mathrm{2}}{\left({y}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\:\mathrm{1}+\frac{\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow \\ $$$$\int\:{f}^{−\mathrm{1}} \left({x}\right){dx}=\:{x}\:+\mathrm{2}\:\int\:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1}}\:+{c}\:{but} \\ $$$$\int\:\:\:\:\frac{\mathrm{2}{dx}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1}}\:=\:\int\:\:\:\frac{\mathrm{2}{dx}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{2}\right)} \\ $$$$=\:\int\:\:\:\:\left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{4}}\right\}{dx}=\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}}\:−\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$$\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}}\:=_{{x}=\sqrt{\mathrm{2}}{t}} \:\:\:\int\:\frac{\sqrt{\mathrm{2}}{dt}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{arctan}\left(\frac{{x}}{\sqrt{\mathrm{2}}}\right) \\ $$$$\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:=_{{x}=\mathrm{2}{t}} \:\:\:\:\int\:\:\frac{\mathrm{2}{dt}}{\mathrm{4}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\frac{{t}}{\mathrm{2}}\right)\Rightarrow \\ $$$$\int\:{f}^{−\mathrm{1}} \left({x}\right){dx}\:=\:{x}\:\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{arctan}\left(\frac{{x}}{\sqrt{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{x}}{\mathrm{2}}\right)+{c} \\ $$

Answered by MJS last updated on 09/Aug/18

∫(√(−3+(√((x+1)/(x−1)))))dx= [t=(√(−3+(√((x+1)/(x−1))))) → dx=−((8t(t^2 +3))/((t^2 +2)^2 (t^2 +4)^2 ))dt] =−8∫((t^2 (t^2 +3))/((t^2 +2)^2 (t^2 +4)^2 ))dt= =−8∫(−(1/(2(t^2 +2)^2 ))+(1/((t^2 +4)^2 ))+(1/(4(t^2 +2)))−(1/(4(t^2 +4))))dt= =4∫(dt/((t^2 +2)^2 ))−8∫(dt/((t^2 +4)^2 ))−2∫(dt/(t^2 +2))+2∫(dt/(t^2 +4))= [∫(dx/((x^2 +p)^2 ))=(x/(2p(x^2 +p)))+((√p^3 )/(2p^3 ))arctan ((x(√p))/p)] [∫(dx/(x^2 +p))=((√p)/p)arctan ((x(√p))/p)] =(t/(t^2 +2))+((√2)/2)arctan ((t(√2))/2) −(t/(t^2 +4))−(1/2)arctan (t/2) −(√2)arctan ((t(√2))/2) +arctan (t/2)= =((2t)/((t^2 +2)(t^2 +4)))+(1/2)(arctan (t/2) −(√2)arctan ((t(√2))/2))= =(√(((√(x+1))−3(√(x−1)))(√((x−1)^3 ))))+(1/2)(arctan((1/2)(√(−3+(√((x+1)/(x−1))))))−(√2)arctan((1/2)(√(−6+2(√((x+1)/(x−1)))))))+C

$$\int\sqrt{−\mathrm{3}+\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{−\mathrm{3}+\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}}\:\rightarrow\:{dx}=−\frac{\mathrm{8}{t}\left({t}^{\mathrm{2}} +\mathrm{3}\right)}{\left({t}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dt}\right] \\ $$$$=−\mathrm{8}\int\frac{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{3}\right)}{\left({t}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dt}= \\ $$$$=−\mathrm{8}\int\left(−\frac{\mathrm{1}}{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}\left({t}^{\mathrm{2}} +\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{4}\left({t}^{\mathrm{2}} +\mathrm{4}\right)}\right){dt}= \\ $$$$=\mathrm{4}\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }−\mathrm{8}\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }−\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}}+\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{4}}= \\ $$$$\:\:\:\:\:\left[\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{p}\right)^{\mathrm{2}} }=\frac{{x}}{\mathrm{2}{p}\left({x}^{\mathrm{2}} +{p}\right)}+\frac{\sqrt{{p}^{\mathrm{3}} }}{\mathrm{2}{p}^{\mathrm{3}} }\mathrm{arctan}\:\frac{{x}\sqrt{{p}}}{{p}}\right] \\ $$$$\:\:\:\:\:\left[\int\frac{{dx}}{{x}^{\mathrm{2}} +{p}}=\frac{\sqrt{{p}}}{{p}}\mathrm{arctan}\:\frac{{x}\sqrt{{p}}}{{p}}\right] \\ $$$$=\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\frac{{t}\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:\frac{{t}}{\mathrm{2}}\:−\sqrt{\mathrm{2}}\mathrm{arctan}\:\frac{{t}\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\mathrm{arctan}\:\frac{{t}}{\mathrm{2}}= \\ $$$$=\frac{\mathrm{2}{t}}{\left({t}^{\mathrm{2}} +\mathrm{2}\right)\left({t}^{\mathrm{2}} +\mathrm{4}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{arctan}\:\frac{{t}}{\mathrm{2}}\:−\sqrt{\mathrm{2}}\mathrm{arctan}\:\frac{{t}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)= \\ $$$$=\sqrt{\left(\sqrt{{x}+\mathrm{1}}−\mathrm{3}\sqrt{{x}−\mathrm{1}}\right)\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\sqrt{−\mathrm{3}+\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}}\right)−\sqrt{\mathrm{2}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\sqrt{−\mathrm{6}+\mathrm{2}\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}}\right)\right)+{C} \\ $$

Answered by ajfour last updated on 10/Aug/18

(y^2 +3)^2 =1+(2/(x−1)) x=1+(2/((y^2 +3)^2 −1)) f^(−1) (x)=1+(2/((x^2 +3)^2 −1)) I=∫[1+(2/((x^2 +3)^2 −1))]dx = x+2I_1 I_1 =∫(dx/((x^2 +3)^2 −1)) = ∫(dx/((x^2 +2)(x^2 +4))) =(1/2)∫(dx/(x^2 +2))−(1/2)∫(dx/(x^2 +4)) hence ∫f^(−1) (x)dx = x+(1/(√2))tan^(−1) (x/(√2)) −(1/2)tan^(−1) (x/2)+c .

$$\left({y}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} =\mathrm{1}+\frac{\mathrm{2}}{{x}−\mathrm{1}} \\ $$$${x}=\mathrm{1}+\frac{\mathrm{2}}{\left({y}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\mathrm{1}+\frac{\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$${I}=\int\left[\mathrm{1}+\frac{\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1}}\right]{dx} \\ $$$$=\:{x}+\mathrm{2}{I}_{\mathrm{1}} \\ $$$${I}_{\mathrm{1}} =\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1}}\:=\:\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$${hence} \\ $$$$\int{f}^{−\mathrm{1}} \left({x}\right){dx}\:=\:{x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}+{c}\:. \\ $$

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