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Question Number 38720 by maxmathsup by imad last updated on 28/Jun/18

find ∫ (((√(x+1)) −(√(x−1)))/((√(x+1)) −(√(x−1))))dx

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

Commented by math khazana by abdo last updated on 28/Jun/18

the Q is find ∫ (((√(x+1)) −(√(x−1)))/((√(x+1)) +(√(x−1))))dx

$${the}\:{Q}\:{is}\:{find}\:\:\int\:\:\:\:\frac{\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}−\mathrm{1}}}{\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}−\mathrm{1}}}{dx} \\ $$

Commented by prof Abdo imad last updated on 29/Jun/18

I= ∫ (({(√(x+1)) −(√(x−1))}^2 )/2)dx = ∫ ((x+1 −2(√(x^2 −1))+x−1)/2)dx =∫ xdx −∫ (√(x^2 −1))dx =(x^2 /2) − ∫ (√(x^2 −1))dx but changement x=ch(t) give ∫(√(x^2 −1))dx=∫ sht shtdt =∫ sh^2 t dt =∫ ((ch(2t)−1)/2)dt =(1/4)sh(2t) −(t/2) =(1/4)2sh(t)ch(t) −(t/2) =(1/2)x(√(x^2 −1)) −(1/2)argch(x) =(x/2)(√(x^2 −1)) −(1/2)ln(x +(√(x^2 −1))) ⇒ I = (x^2 /2) −(x/2)(√(x^2 −1)) +(1/2)ln(x+(√(x^2 −1))) +c

$${I}=\:\int\:\:\:\frac{\left\{\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}−\mathrm{1}}\right\}^{\mathrm{2}} }{\mathrm{2}}{dx} \\ $$$$=\:\int\:\:\:\:\frac{{x}+\mathrm{1}\:−\mathrm{2}\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}+{x}−\mathrm{1}}{\mathrm{2}}{dx} \\ $$$$=\int\:\:\:{xdx}\:−\int\:\:\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:−\:\int\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx}\:{but}\:{changement}\:{x}={ch}\left({t}\right) \\ $$$${give}\:\int\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx}=\int\:\:{sht}\:{shtdt} \\ $$$$=\int\:{sh}^{\mathrm{2}} {t}\:{dt}\:=\int\:\:\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)\:−\frac{{t}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{2}{sh}\left({t}\right){ch}\left({t}\right)\:−\frac{{t}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}{argch}\left({x}\right) \\ $$$$=\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow \\ $$$${I}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\:−\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:+{c} \\ $$$$ \\ $$

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