Question Number 44498 by peter frank last updated on 30/Sep/18
Commented by maxmathsup by imad last updated on 30/Sep/18
$${changement}\:\mathrm{4}^{{x}} ={t}\:{give}\:\:{e}^{{xln}\left(\mathrm{4}\right)} ={t}\:\Rightarrow{xln}\left(\mathrm{4}\right)={ln}\left({t}\right)\:\Rightarrow{dx}=\frac{\mathrm{1}}{{tln}\left(\mathrm{4}\right)}{dt} \\ $$$${I}\:=\:\int\mathrm{4}^{\mathrm{4}^{{t}} } \:.\mathrm{4}^{{t}} \:{t}\:\:\frac{{dt}}{{t}\:{ln}\left(\mathrm{4}\right)}\:=\frac{\mathrm{1}}{{ln}\left(\mathrm{4}\right)}\:\int\:\:\mathrm{4}^{{t}} .\:\mathrm{4}^{\mathrm{4}^{{t}} } \:\:{dt}\:\:\:{also}\:{changement}\:\:\mathrm{4}^{{t}} ={u}\:{give} \\ $$$${dt}\:=\frac{{du}}{{u}\:{ln}\left(\mathrm{4}\right)}\:\Rightarrow\:{I}\:=\:\frac{\mathrm{1}}{{ln}\left(\mathrm{4}\right)}\int\:\:{u}\:.\mathrm{4}^{{u}} \:\frac{{du}}{{u}\:{ln}\left(\mathrm{4}\right)}\:=\frac{\mathrm{1}}{{ln}\left(\mathrm{4}\right)}\:\int\:\:\mathrm{4}^{{u}} \:{du} \\ $$$$=\frac{\mathrm{1}}{{ln}\left(\mathrm{4}\right)^{\mathrm{2}} }\:\int\:\:\:{e}^{{uln}\left(\mathrm{4}\right)} {du}\:=\frac{\mathrm{1}}{\left({ln}\left(\mathrm{4}\right)\right)^{\mathrm{3}} }\:\mathrm{4}^{{u}} \:\:\:=\frac{\mathrm{1}}{\left({ln}\left(\mathrm{4}\right)\right)^{\mathrm{3}} }\:\mathrm{4}^{\mathrm{4}^{{t}} } \:=\frac{\mathrm{1}}{\left({ln}\left(\mathrm{4}\right)\right)^{\mathrm{3}} }\:\mathrm{4}^{\mathrm{4}^{\mathrm{4}^{{x}} } } \:+{c} \\ $$$${perhaps}\:{there}\:{is}\:{something}\:{wrong}\:{in}\:{the}\:{question}..! \\ $$
Commented by peter frank last updated on 30/Sep/18
$$\:\mathrm{true}\:\mathrm{sir}\:\mathrm{its}\:\mathrm{typing}\:\mathrm{error}\:\frac{}{\left(\mathrm{log}_{\mathrm{e}\:} \mathrm{4}\right)^{\mathrm{3}} }\:\:\:\:\:\: \\ $$