Question Number 185848 by Mingma last updated on 28/Jan/23
Answered by Frix last updated on 28/Jan/23
$$\int\sqrt{\mathrm{e}^{{x}} +\mathrm{2}}{dx}\:\overset{{t}=\sqrt{\mathrm{e}^{{x}} +\mathrm{2}}} {=}\:\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{2}}{dt}= \\ $$$$=\mathrm{2}\int{dt}+\sqrt{\mathrm{2}}\int\frac{{dt}}{{t}−\sqrt{\mathrm{2}}}−\sqrt{\mathrm{2}}\int\frac{{dt}}{{t}+\sqrt{\mathrm{2}}}= \\ $$$$=\mathrm{2}{t}+\sqrt{\mathrm{2}}\mathrm{ln}\:\left({t}−\sqrt{\mathrm{2}}\right)\:−\sqrt{\mathrm{2}}\mathrm{ln}\:\left({t}+\sqrt{\mathrm{2}}\right)\:= \\ $$$$… \\ $$$$=\mathrm{2}\sqrt{\mathrm{e}^{{x}} +\mathrm{2}}−\sqrt{\mathrm{2}}{x}+\mathrm{2}\sqrt{\mathrm{2}}\mathrm{ln}\:\left(\sqrt{\mathrm{e}^{{x}} +\mathrm{2}}−\sqrt{\mathrm{2}}\right)\:+{C} \\ $$$$ \\ $$