Question Number 146215 by mathdanisur last updated on 12/Jul/21
$$\int\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 12/Jul/21
$$\mathrm{F}\left({x}\right)\:=\:\int\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}} \\ $$$$\mathrm{F}\left(\mathrm{2ch}{u}\right)\:=\:\int\frac{\mathrm{2sh}{udu}}{\:\mathrm{2ch}{u}\sqrt{\mathrm{4ch}^{\mathrm{2}} {u}−\mathrm{4}}} \\ $$$$\mathrm{F}\left(\mathrm{2ch}{u}\right)\:=\:\int\frac{\mathrm{sh}{udu}}{\:\mathrm{ch}{u}×\mathrm{2sh}{u}} \\ $$$$\mathrm{F}\left(\mathrm{2ch}{u}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{\:\mathrm{ch}{u}} \\ $$$$\mathrm{F}\left(\mathrm{2ch}{u}\right)\:=\:\int\frac{{e}^{{u}} {du}}{\:{e}^{\mathrm{2}{u}} +\mathrm{1}} \\ $$$$\mathrm{F}\left(\mathrm{2ch}{u}\right)\:=\:\mathrm{arctan}\left({e}^{{u}} \right)+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{arctan}\left({e}^{\mathrm{argch}\frac{{x}}{\mathrm{2}}} \right)+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{arctan}\left({e}^{\mathrm{ln}\left(\frac{{x}}{\mathrm{2}}+\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}}\right)} \right)+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{arctan}\left(\frac{{x}}{\mathrm{2}}+\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}}\right)+\mathrm{C} \\ $$
Answered by qaz last updated on 12/Jul/21
$$\int\frac{\mathrm{dx}}{\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{4}}}=\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }}}=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{d}\left(\frac{\mathrm{2}}{\mathrm{x}}\right)}{\:\sqrt{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }}}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{x}}+\mathrm{C} \\ $$
Answered by mathmax by abdo last updated on 12/Jul/21
$$\Phi=\int\:\:\frac{\mathrm{dx}}{\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{4}}}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}=\frac{\mathrm{2}}{\mathrm{sint}}\:\Rightarrow\frac{\mathrm{dx}}{\mathrm{dt}}=−\frac{\mathrm{2cost}}{\mathrm{sin}^{\mathrm{2}} \mathrm{t}} \\ $$$$\Rightarrow\Phi=−\mathrm{2}\int\:\:\frac{\mathrm{cost}}{\mathrm{sin}^{\mathrm{2}} \mathrm{t}×\frac{\mathrm{2}}{\mathrm{sint}}\sqrt{\frac{\mathrm{4}}{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}−\mathrm{4}}}\mathrm{dt} \\ $$$$=−\int\:\:\frac{\mathrm{cost}}{\mathrm{sint}×\mathrm{2}×\frac{\mathrm{cost}}{\mathrm{sint}}}\mathrm{dt}=−\frac{\mathrm{1}}{\mathrm{2}}\int\:\mathrm{dt}\:=−\frac{\mathrm{t}}{\mathrm{2}}\:\mathrm{C} \\ $$$$\mathrm{sint}=\frac{\mathrm{2}}{\mathrm{x}}\:\Rightarrow\mathrm{t}=\mathrm{arcsin}\left(\frac{\mathrm{2}}{\mathrm{x}}\right)\:\Rightarrow\Phi=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arcsin}\left(\frac{\mathrm{2}}{\mathrm{x}}\right)\:+\mathrm{C} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 12/Jul/21
$${Thanks}\:{Ser} \\ $$