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Question Number 137004 by Mathspace last updated on 28/Mar/21
find ∫ ((√x)/( (√(x−1))+(√(x+1))))dx
$${find}\:\int\:\frac{\sqrt{{x}}}{\:\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{1}}}{dx} \\ $$
Answered by aleks041103 last updated on 28/Mar/21
((√x)/( (√(x−1))+(√(x+1))))= =((√x)/( (√(x−1))+(√(x+1)))) (((√(x+1))−(√(x−1)))/( (√(x+1))−(√(x−1))))= =(√x)((√(x+1))−(√(x−1))) I(a)=∫(√(x(x+2a)))dx = =∫(√(((x+a)−a)((x+a)+a)))dx= =∫(√((x+a)^2 −a^2 ))dx= =a^2 ∫(√((1+x/a)^2 −1)) d(1+(x/a)) cosh(t) = 1+(x/a) sinh(t)dt = d(1+(x/a)) I(a)=a^2 ∫(√(cosh^2 (t)−1))sinh(t)dt= =a^2 ∫sinh^2 (t)dt sinh^2 (t)=cosh(2t)−1 I(a)=a^2 ((1/2)sinh(2t)−t) t = arccosh(1+x/a)=ln(1+(x/a)+(√((1+x/a)^2 −1)))= =ln(1+a^(−1) x+a^(−1) (√(x(x+2a)))) (1/2)sinh(2t)=sinh(t)cosh(t)= =cosh(t)(√(cosh^2 (t)−1))= =(1+x/a)(√((1+x/a)^2 −1))= =a^(−2) (x+a)(√(x(x+2a))) I(a)=(x+a)(√(x(x+2a)))−a^2 ln(1+a^(−1) x+a^(−1) (√(x(x+2a)))) ∫ ((√x)/( (√(x−1))+(√(x+1))))dx=∫ (√x)((√(x+1))−(√(x−1))) dx= =I(1/2)−I(−1/2)= =(1/2)((2x+1)(√(x(x+1)))−(2x−1)(√(x(x−1))))−(1/4)ln(((1+2x+2(√(x(x+1))))/(1−2x−2(√(x(x−1)))))) ∫ ((√x)/( (√(x−1))+(√(x+1))))dx = =((2x+1)(√(x(x+1)))−(2x−1)(√(x(x−1))))−(1/4)ln(((1+2x+2(√(x(x+1))))/(1−2x−2(√(x(x−1))))))+C
$$\frac{\sqrt{{x}}}{\:\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{1}}}= \\ $$$$=\frac{\sqrt{{x}}}{\:\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{1}}}\:\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}}= \\ $$$$=\sqrt{{x}}\left(\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}\right) \\ $$$$ \\ $$$${I}\left({a}\right)=\int\sqrt{{x}\left({x}+\mathrm{2}{a}\right)}{dx}\:=\: \\ $$$$=\int\sqrt{\left(\left({x}+{a}\right)−{a}\right)\left(\left({x}+{a}\right)+{a}\right)}{dx}= \\ $$$$=\int\sqrt{\left({x}+{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }{dx}= \\ $$$$={a}^{\mathrm{2}} \int\sqrt{\left(\mathrm{1}+{x}/{a}\right)^{\mathrm{2}} −\mathrm{1}}\:{d}\left(\mathrm{1}+\frac{{x}}{{a}}\right) \\ $$$${cosh}\left({t}\right)\:=\:\mathrm{1}+\frac{{x}}{{a}} \\ $$$${sinh}\left({t}\right){dt}\:=\:{d}\left(\mathrm{1}+\frac{{x}}{{a}}\right) \\ $$$${I}\left({a}\right)={a}^{\mathrm{2}} \int\sqrt{{cosh}^{\mathrm{2}} \left({t}\right)−\mathrm{1}}{sinh}\left({t}\right){dt}= \\ $$$$={a}^{\mathrm{2}} \int{sinh}^{\mathrm{2}} \left({t}\right){dt} \\ $$$${sinh}^{\mathrm{2}} \left({t}\right)={cosh}\left(\mathrm{2}{t}\right)−\mathrm{1} \\ $$$${I}\left({a}\right)={a}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{sinh}\left(\mathrm{2}{t}\right)−{t}\right) \\ $$$${t}\:=\:{arccosh}\left(\mathrm{1}+{x}/{a}\right)={ln}\left(\mathrm{1}+\frac{{x}}{{a}}+\sqrt{\left(\mathrm{1}+{x}/{a}\right)^{\mathrm{2}} −\mathrm{1}}\right)= \\ $$$$={ln}\left(\mathrm{1}+{a}^{−\mathrm{1}} {x}+{a}^{−\mathrm{1}} \sqrt{{x}\left({x}+\mathrm{2}{a}\right)}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{sinh}\left(\mathrm{2}{t}\right)={sinh}\left({t}\right){cosh}\left({t}\right)= \\ $$$$={cosh}\left({t}\right)\sqrt{{cosh}^{\mathrm{2}} \left({t}\right)−\mathrm{1}}= \\ $$$$=\left(\mathrm{1}+{x}/{a}\right)\sqrt{\left(\mathrm{1}+{x}/{a}\right)^{\mathrm{2}} −\mathrm{1}}= \\ $$$$={a}^{−\mathrm{2}} \left({x}+{a}\right)\sqrt{{x}\left({x}+\mathrm{2}{a}\right)} \\ $$$${I}\left({a}\right)=\left({x}+{a}\right)\sqrt{{x}\left({x}+\mathrm{2}{a}\right)}−{a}^{\mathrm{2}} {ln}\left(\mathrm{1}+{a}^{−\mathrm{1}} {x}+{a}^{−\mathrm{1}} \sqrt{{x}\left({x}+\mathrm{2}{a}\right)}\right) \\ $$$$ \\ $$$$\int\:\frac{\sqrt{{x}}}{\:\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{1}}}{dx}=\int\:\sqrt{{x}}\left(\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}\right)\:{dx}= \\ $$$$={I}\left(\mathrm{1}/\mathrm{2}\right)−{I}\left(−\mathrm{1}/\mathrm{2}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}\left({x}+\mathrm{1}\right)}−\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{{x}\left({x}−\mathrm{1}\right)}\right)−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\frac{\mathrm{1}+\mathrm{2}{x}+\mathrm{2}\sqrt{{x}\left({x}+\mathrm{1}\right)}}{\mathrm{1}−\mathrm{2}{x}−\mathrm{2}\sqrt{{x}\left({x}−\mathrm{1}\right)}}\right) \\ $$$$\int\:\frac{\sqrt{{x}}}{\:\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{1}}}{dx}\:= \\ $$$$=\left(\left(\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}\left({x}+\mathrm{1}\right)}−\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{{x}\left({x}−\mathrm{1}\right)}\right)−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\frac{\mathrm{1}+\mathrm{2}{x}+\mathrm{2}\sqrt{{x}\left({x}+\mathrm{1}\right)}}{\mathrm{1}−\mathrm{2}{x}−\mathrm{2}\sqrt{{x}\left({x}−\mathrm{1}\right)}}\right)+{C} \\ $$
Answered by mathmax by abdo last updated on 29/Mar/21
thankx sir.
$$\mathrm{thankx}\:\mathrm{sir}. \\ $$
Answered by EDWIN88 last updated on 29/Mar/21
((√x)/( (√(x−1)) +(√(x+1)))) = (((√x) ((√(x−1)) −(√(x+1))))/((x−1)−(x+1))) =−(((√x) ((√(x−1)) −(√(x+1)) ))/2) E = −(1/2)∫(√(x^2 −x)) dx +(1/2)∫ (√(x^2 +x)) dx E=−(1/2)∫ (√((x−(1/2))^2 −(1/4))) dx +(1/2)∫ (√((x+(1/2))^2 −(1/4))) dx E_1 =−(1/2)∫ (√((x−(1/2))^2 −(1/4))) dx let x−(1/2) = (1/2)sec t dx =(1/2)sec t tan t dt E_1 =−(1/2)∫ (1/2)tan t .(1/2)sec t tan t dt E_1 =−(1/8)∫ sec t (sec^2 t−1) dt E_1 =−(1/8)∫ sec^3 t−sec t dt similary to E_2 =(1/2)∫ (√((x+(1/2))^2 −(1/4))) dx
$$\frac{\sqrt{\mathrm{x}}}{\:\sqrt{\mathrm{x}−\mathrm{1}}\:+\sqrt{\mathrm{x}+\mathrm{1}}}\:=\:\frac{\sqrt{\mathrm{x}}\:\left(\sqrt{\mathrm{x}−\mathrm{1}}\:−\sqrt{\mathrm{x}+\mathrm{1}}\right)}{\left(\mathrm{x}−\mathrm{1}\right)−\left(\mathrm{x}+\mathrm{1}\right)} \\ $$$$=−\frac{\sqrt{\mathrm{x}}\:\left(\sqrt{\mathrm{x}−\mathrm{1}}\:−\sqrt{\mathrm{x}+\mathrm{1}}\:\right)}{\mathrm{2}} \\ $$$$\mathrm{E}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{x}}\:\mathrm{dx}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}\:\mathrm{dx} \\ $$$$\mathrm{E}=−\frac{\mathrm{1}}{\mathrm{2}}\int\:\sqrt{\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}\:\mathrm{dx}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\:\sqrt{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}\:\mathrm{dx} \\ $$$$\mathrm{E}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\int\:\sqrt{\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}\:\mathrm{dx} \\ $$$$\mathrm{let}\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:\mathrm{t}\: \mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:\mathrm{t}\:\mathrm{tan}\:\mathrm{t}\:\mathrm{dt} \\ $$$$\mathrm{E}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\mathrm{t}\:.\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:\mathrm{t}\:\mathrm{tan}\:\mathrm{t}\:\mathrm{dt} \\ $$$$\mathrm{E}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{8}}\int\:\mathrm{sec}\:\mathrm{t}\:\left(\mathrm{sec}\:^{\mathrm{2}} \:\mathrm{t}−\mathrm{1}\right)\:\mathrm{dt}\: \\ $$$$\mathrm{E}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{8}}\int\:\mathrm{sec}\:^{\mathrm{3}} \mathrm{t}−\mathrm{sec}\:\mathrm{t}\:\mathrm{dt}\: \\ $$$$\mathrm{similary}\:\mathrm{to}\:\mathrm{E}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\int\:\sqrt{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}\:\mathrm{dx} \\ $$

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