Question Number 179999 by yaslm last updated on 05/Nov/22
Commented by mokys last updated on 05/Nov/22
$${lnx}\:−\:\mathrm{1}\:=\:{y}\:\rightarrow\:{lnx}\:=\:{y}\:+\:\mathrm{1}\rightarrow\:{x}\:=\:{e}^{{y}+\mathrm{1}} \rightarrow\:{dx}\:=\:{e}^{{y}+\mathrm{1}} {dy} \\ $$$$ \\ $$$${I}\:=\:\int\:\left({y}+\mathrm{1}\right)\:{e}^{{y}^{\mathrm{2}} +{y}} \:{e}^{{y}+\mathrm{1}} {dy}\:=\:\int\:\left({y}+\mathrm{1}\right)\:{e}^{\left({y}+\mathrm{1}\right)^{\mathrm{2}} } {dy}\: \\ $$$$ \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{\left({y}+\mathrm{1}\right)^{\mathrm{2}} } \:+\:{C}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{ln}^{\mathrm{2}} {x}} \:+{c} \\ $$
Answered by Gamil last updated on 05/Nov/22
