Question Number 38720 by maxmathsup by imad last updated on 28/Jun/18
$${find}\:\:\:\int\:\:\:\:\:\frac{\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}−\mathrm{1}}}{dx} \\ $$
Commented by math khazana by abdo last updated on 28/Jun/18
$${the}\:{Q}\:{is}\:{find}\:\:\int\:\:\:\:\frac{\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}−\mathrm{1}}}{dx} \\ $$
Commented by prof Abdo imad last updated on 29/Jun/18
$${I}=\:\int\:\:\:\frac{\left\{\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}−\mathrm{1}}\right\}^{\mathrm{2}} }{\mathrm{2}}{dx} \\ $$$$=\:\int\:\:\:\:\frac{{x}+\mathrm{1}\:−\mathrm{2}\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}+{x}−\mathrm{1}}{\mathrm{2}}{dx} \\ $$$$=\int\:\:\:{xdx}\:−\int\:\:\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:−\:\int\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx}\:{but}\:{changement}\:{x}={ch}\left({t}\right) \\ $$$${give}\:\int\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx}=\int\:\:{sht}\:{shtdt} \\ $$$$=\int\:{sh}^{\mathrm{2}} {t}\:{dt}\:=\int\:\:\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)\:−\frac{{t}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{2}{sh}\left({t}\right){ch}\left({t}\right)\:−\frac{{t}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}{argch}\left({x}\right) \\ $$$$=\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow \\ $$$${I}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\:−\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:+{c} \\ $$$$ \\ $$