Question Number 73482 by abdomathmax last updated on 13/Nov/19
$${find}\:\int\:\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}} \\ $$
Commented by abdomathmax last updated on 17/Nov/19
$${let}\:{I}\:=\int\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}}\:\Rightarrow \\ $$$${I}\:=\:\int\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$${we}\:{have}\:\int\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}{dx}\:=_{{x}=\sqrt{\mathrm{3}}{sh}\left({t}\right)} \:\sqrt{\mathrm{3}}\:\int\:\sqrt{\mathrm{3}}{ch}\left({t}\right){ch}\left({t}\right) \\ $$$$=\mathrm{3}\int\:\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt}\:=\frac{\mathrm{3}}{\mathrm{2}}{t}\:+\frac{\mathrm{3}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)\:+{c}_{\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{t}\:\:+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{2}{sh}\left({t}\right){ch}\left({t}\right)\:+{c}_{\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\:{argsh}\left(\frac{{x}}{\:\sqrt{\mathrm{3}}}\right)\:+\frac{\mathrm{3}}{\mathrm{2}}\frac{{x}}{\:\sqrt{\mathrm{3}}}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}\:+{c} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{ln}\left(\frac{{x}}{\:\sqrt{\mathrm{3}}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}\right)\:+\frac{{x}}{\mathrm{2}}\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }\:+{c}_{\mathrm{1}} \\ $$$$\int\:\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\:=_{{x}={sh}\left({t}\right)} \:\:\:\int{ch}\left({t}\right){ch}\left({t}\right){dt} \\ $$$$=\int\:\:\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt}\:=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)\:+{c}_{\mathrm{2}} \\ $$$$=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({g}\right){ch}\left({t}\right)\:+{c}_{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\frac{{x}}{\mathrm{2}}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+{c}_{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{3}}{\mathrm{8}}{ln}\left({x}+\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }\right)\:+\frac{{x}}{\mathrm{4}}\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right) \\ $$$$−\frac{{x}}{\mathrm{4}}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+{C} \\ $$$$ \\ $$
Answered by ajfour last updated on 13/Nov/19
$${I}=\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\right){dx} \\ $$$$\:\:\:=\frac{{x}}{\mathrm{4}}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\right) \\ $$$$+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ln}\:\mid{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\mid−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\mid+{c} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_. \\ $$
Commented by abdomathmax last updated on 17/Nov/19
$${thanks}\:{sir}. \\ $$