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Question Number 78655 by Pratah last updated on 19/Jan/20

Commented by mr W last updated on 19/Jan/20

$$=e^{e^{..{.}^{e^e}}}-e^{e^{..{.}^e}}$$

Commented by john santu last updated on 19/Jan/20

$$hahaha....$$

Answered by mind is power last updated on 19/Jan/20

$$\int {\mathrm{e}}^{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}={\mathrm{e}}^{{\mathrm{e}}^{\mathrm{x}}}+\mathrm{c}$$$$\int {\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{\mathrm{x}}}}.{\mathrm{e}}^{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx},{\mathrm{e}}^{\mathrm{x}}=\mathrm{u}=\int {\mathrm{e}}^{{\mathrm{e}}^{\mathrm{u}}}{\mathrm{e}}^{\mathrm{u}}\mathrm{du}={\mathrm{e}}^{{\mathrm{e}}^{\mathrm{u}}}+\mathrm{c}={\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{\mathrm{x}}}}+\mathrm{c}$$$$\mathrm{let}{\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)={\mathrm{e}}^{....{\mathrm{e}}^{\mathrm{x}}},\mathrm{n}\mathrm{times}$$$${\mathrm{f}}_1\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}},{\mathrm{f}}_2\left(\mathrm{x}\right)={\mathrm{e}}^{{\mathrm{e}}^{\mathrm{x}}},\mathrm{claim}$$$$\int {\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right).{\mathrm{f}}_{\mathrm{n}-1}\left(\mathrm{x}\right).....{\mathrm{f}}_1\left(\mathrm{x}\right)\mathrm{dx}={\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)+\mathrm{c}$$$$\mathrm{By}\mathrm{inductition}\mathrm{n}=1\int {\mathrm{f}}_1\left(\mathrm{x}\right)\mathrm{dx}=\int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}={\mathrm{e}}^{\mathrm{x}}+\mathrm{c}={\mathrm{f}}_1+\mathrm{c}\mathrm{True}$$$$\int {\mathrm{f}}_{\mathrm{n}+1}\left(\mathrm{x}\right).{\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)......{\mathrm{f}}_1\left(\mathrm{x}\right)\mathrm{dx}=\int {\mathrm{f}}_{\mathrm{n}+1}\left(\mathrm{x}\right)...{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}$$$${\mathrm{e}}^{\mathrm{x}}=\mathrm{u}\Rightarrow {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}=\mathrm{du}$$$$\int {\mathrm{f}}_{\mathrm{n}+1}\left(\mathrm{x}\right).{\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)......{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}=\int {\mathrm{f}}_{\mathrm{n}}\left({\mathrm{e}}^{\mathrm{x}}\right).....{\mathrm{f}}_1\left({\mathrm{e}}^{\mathrm{x}}\right).{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}$$$$=\int {\mathrm{f}}_{\mathrm{n}}\left(\mathrm{u}\right).....{\mathrm{f}}_1\left(\mathrm{u}\right)\mathrm{du}={\mathrm{f}}_{\mathrm{n}}\left(\mathrm{u}\right)+\mathrm{c}\mathrm{By}\mathrm{hypothese}\mathrm{of}\mathrm{recursion}$$$$={\mathrm{f}}_{\mathrm{n}}\left({\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{c}={\mathrm{f}}_{\mathrm{n}+1}\left(\mathrm{x}\right)+\mathrm{c}\mathrm{Proved}$$$$\mathrm{in}\mathrm{our}\mathrm{exemple}\mathrm{n}=10$$$$\int {\mathrm{f}}_{10}\left(\mathrm{x}\right)......{\mathrm{f}}_1\left(\mathrm{x}\right)\mathrm{dx}={\mathrm{f}}_{10}\left(\mathrm{x}\right)+\mathrm{c}$$$$$$$$$$$$$$

Commented by Pratah last updated on 19/Jan/20

$$\mathrm{thanks}$$

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