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Question Number 95115 by bobhans last updated on 23/May/20

$$\int {\mathrm{e}}^x\sqrt{1+e^{2x}}dx=?$$

Answered by i jagooll last updated on 23/May/20

Commented by bobhans last updated on 23/May/20

$$\mathrm{thank}\mathrm{you}$$

Commented by peter frank last updated on 23/May/20

$$\mathrm{thank}\mathrm{you}$$

Answered by mathmax by abdo last updated on 23/May/20

$$\mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}\sqrt{1+{\mathrm{e}}^{2\mathrm{x}}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\sqrt{1+{\mathrm{e}}^{2\mathrm{x}}}=\mathrm{t}\Rightarrow$$$$1+{\mathrm{e}}^{2\mathrm{x}}={\mathrm{t}}^2\Rightarrow {\mathrm{e}}^{2\mathrm{x}}={\mathrm{t}}^2-1\Rightarrow \mathrm{x}=\frac{1}{2}\ln ({\mathrm{t}}^2-1)\Rightarrow \mathrm{dx}=\frac{\mathrm{t}}{{\mathrm{t}}^2-1}\mathrm{dt}\Rightarrow$$$$\mathrm{I}=\int \sqrt{{\mathrm{t}}^2-1}\times \mathrm{t}\times \frac{\mathrm{t}}{{\mathrm{t}}^2-1}\mathrm{dt}=\int \frac{{\mathrm{t}}^2}{\sqrt{{\mathrm{t}}^2-1}}\mathrm{dt}$$$$=\int \frac{{\mathrm{t}}^2-1+1}{\sqrt{{\mathrm{t}}^2-1}}\mathrm{dt}=\int \sqrt{{\mathrm{t}}^2-1}\mathrm{dt}+\int \frac{\mathrm{dt}}{\sqrt{{\mathrm{t}}^2-1}}$$$$\mathrm{changement}\mathrm{t}=\mathrm{chu}\mathrm{give}$$$$\int \sqrt{{\mathrm{t}}^2-1}\mathrm{dt}=\int {\mathrm{sh}}^2\mathrm{u}\mathrm{du}=\int \frac{\mathrm{ch}\left(2\mathrm{u}\right)-1}{2}\mathrm{du}=\frac{1}{4}\mathrm{sh}\left(2\mathrm{u}\right)-\frac{\mathrm{u}}{2}+{\mathrm{c}}_0$$$$=\frac{1}{2}\mathrm{sh}\left(\mathrm{u}\right)\mathrm{chu}-\frac{\mathrm{u}}{2}+{\mathrm{c}}_0=\frac{1}{2}\mathrm{t}\sqrt{{\mathrm{t}}^2-1}-\frac{1}{2}\ln (\mathrm{t}+\sqrt{{\mathrm{t}}^2-1})+{\mathrm{c}}_0$$$$\int \frac{\mathrm{dt}}{\sqrt{{\mathrm{t}}^2-1}}{=}_{\mathrm{t}=\mathrm{chu}}\int \frac{\mathrm{shu}}{\mathrm{shu}}\mathrm{du}=\mathrm{u}+{\mathrm{c}}_1=\ln (\mathrm{t}+\sqrt{{\mathrm{t}}^2-1})+{\mathrm{c}}_1\Rightarrow$$$$\mathrm{I}=\frac{\mathrm{t}\sqrt{{\mathrm{t}}^2-1}}{2}+\frac{1}{2}\ln (\mathrm{t}+\sqrt{{\mathrm{t}}^2-1})+\mathrm{C}$$$$=\frac{1}{2}\sqrt{1+{\mathrm{e}}^{2\mathrm{x}}}\times \sqrt{1+{\mathrm{e}}^{2\mathrm{x}}-1}+\frac{1}{2}\ln (\sqrt{1+{\mathrm{e}}^{2\mathrm{x}}}+\sqrt{1+{\mathrm{e}}^{2\mathrm{x}}-1})+\mathrm{C}$$$$=\frac{{\mathrm{e}}^{\mathrm{x}}\sqrt{1+{\mathrm{e}}^{2\mathrm{x}}}}{2}+\frac{1}{2}\ln (\mathrm{x}+\sqrt{1+{\mathrm{e}}^{2\mathrm{x}}})+\mathrm{C}$$

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