Question Number 73545 by Rio Michael last updated on 13/Nov/19
$${evaluate}\:\:\int{lnx}\:{dx} \\ $$
Commented by Tony Lin last updated on 13/Nov/19
$${integration}\:{by}\:{part} \\ $$$$\int{f}\:'\left({x}\right){g}\left({x}\right)={f}\left({x}\right){g}\left({x}\right)−\int{f}\left({x}\right){g}\:'\left({x}\right) \\ $$$$\int{lnxdx} \\ $$$$=\int\left(\mathrm{1}×{lnx}\right){dx} \\ $$$$={xlnx}−\int\left({x}×\frac{\mathrm{1}}{{x}}\right){dx} \\ $$$$={xlnx}−{x}+{c} \\ $$
Commented by Rio Michael last updated on 13/Nov/19
$${thanks} \\ $$
Answered by ajfour last updated on 13/Nov/19
$$\frac{{d}}{{dx}}\left({x}\mathrm{ln}\:{x}\right)=\mathrm{ln}\:{x}+\mathrm{1} \\ $$$$\Rightarrow\:{d}\left({x}\mathrm{ln}\:{x}\right)=\:\left(\mathrm{ln}\:{x}\right){dx}+{dx} \\ $$$${Integrating} \\ $$$$\:\:\int{d}\left({x}\mathrm{ln}\:{x}\right)=\int\left(\mathrm{ln}\:{x}\right){dx}+\int{dx} \\ $$$$\:\:{x}\mathrm{ln}\:{x}+{c}\:=\:\int\left(\mathrm{ln}\:{x}\right){dx}\:+\:{x} \\ $$$$\Rightarrow\:\:\int\mathrm{ln}\:{x}\:{dx}\:=\:{x}\mathrm{ln}\:{x}−{x}+{c}\:. \\ $$$$ \\ $$
Commented by Rio Michael last updated on 13/Nov/19
$${i}\:{appreciate}\:{sir} \\ $$