Question Number 69597 by ozodbek last updated on 25/Sep/19
Commented by mathmax by abdo last updated on 25/Sep/19
$${generally}\:{let}\:{find}\:{f}\left({a}\right)\:=\int\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}\:\:{with}\:{a}>\mathrm{0} \\ $$$${changement}\:{x}\:={ash}\left({t}\right)\:{give}\:{f}\left({a}\right)=\int{ach}\left({t}\right){ach}\left({t}\right){dt} \\ $$$$={a}^{\mathrm{2}} \int\:{ch}^{\mathrm{2}} \left({t}\right){dt}\:={a}^{\mathrm{2}} \int\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt}\:=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{t}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\:\int{ch}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{t}\:\:+\:\frac{{a}^{\mathrm{2}} }{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)+{c}\:=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{t}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{sh}\left({t}\right){ch}\left({t}\right)\:+{c} \\ $$$${we}\:{have}\:{t}\:={argsh}\left(\frac{{x}}{{a}}\right)\:={ln}\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)\:={ln}\left(\frac{{x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{{a}}\right)\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{ln}\left(\frac{{x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{{a}}\right)+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\frac{{x}}{{a}}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:+{c} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right)−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{ln}\left({a}\right)\:+\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:+{c} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right)\:+\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:+{C} \\ $$$${a}=\mathrm{2}\:\Rightarrow\:\int\:\sqrt{\mathrm{4}+{x}^{\mathrm{2}} }{dx}\:=\mathrm{2}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\right)+\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:+{C} \\ $$
Answered by MJS last updated on 25/Sep/19
$$\int\sqrt{\mathrm{4}+{x}^{\mathrm{2}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{sinh}^{−\mathrm{1}} \:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}{dt}\right] \\ $$$$=\mathrm{4}\int\mathrm{cosh}^{\mathrm{2}} \:{t}\:{dt}=\mathrm{2}\int{dt}+\mathrm{2}\int\mathrm{cosh}\:\mathrm{2}{t}\:{dt}= \\ $$$$=\mathrm{2}{t}+\mathrm{sinh}\:\mathrm{2}{t}\:=\mathrm{2sinh}^{−\mathrm{1}} \:\frac{{x}}{\mathrm{2}}\:+\frac{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\:+{C} \\ $$$$\mathrm{different}\:\mathrm{path}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{2ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\right)\:+\frac{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\:+{C} \\ $$