Question Number 65354 by mathmax by abdo last updated on 28/Jul/19
$${find}\:\:\int\:\:\:\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{2}}} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 29/Jul/19
$${let}\:{A}\:=\int\:\:\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+{x}−\mathrm{2}}}\:\:{we}\:{have}\:\:{x}^{\mathrm{2}} +{x}−\mathrm{2}\:={x}^{\mathrm{2}} −\mathrm{1}\:+{x}−\mathrm{1} \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)+{x}−\mathrm{1}\:=\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right)\:\Rightarrow{A}\:=\int\:\:\:\frac{{dx}}{\:\sqrt{{x}−\mathrm{1}}\sqrt{{x}−\mathrm{2}}} \\ $$$${changement}\:\sqrt{{x}−\mathrm{1}}={t}\:{give}\:{x}−\mathrm{1}\:={t}^{\mathrm{2}} \:\Rightarrow{x}\:={t}^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow \\ $$$${A}\:=\int\:\:\:\:\frac{\mathrm{2}{tdt}}{{t}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}\:\:=\mathrm{2}\:\int\:\frac{{dt}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}\:=\mathrm{2}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)\:+{c} \\ $$$$=\mathrm{2}{ln}\left(\sqrt{{x}−\mathrm{1}}+\sqrt{{x}−\mathrm{2}}\right)\:+{c}\:\: \\ $$
Commented by MJS last updated on 29/Jul/19
$$\mathrm{typo}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end} \\ $$$$\mathrm{2ln}\:\left(\sqrt{{x}−\mathrm{1}}+\sqrt{{x}+\mathrm{2}}\right)\:+{C} \\ $$
Commented by mathmax by abdo last updated on 30/Jul/19
$${thank}\:{you}\:{sir}\:{MJS} \\ $$