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Question-185848




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

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