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Question Number 73545 by Rio Michael last updated on 13/Nov/19

evaluate  ∫lnx dx

$${evaluate}\:\:\int{lnx}\:{dx} \\ $$

Commented by Tony Lin last updated on 13/Nov/19

integration by part  ∫f ′(x)g(x)=f(x)g(x)−∫f(x)g ′(x)  ∫lnxdx  =∫(1×lnx)dx  =xlnx−∫(x×(1/x))dx  =xlnx−x+c

$${integration}\:{by}\:{part} \\ $$$$\int{f}\:'\left({x}\right){g}\left({x}\right)={f}\left({x}\right){g}\left({x}\right)−\int{f}\left({x}\right){g}\:'\left({x}\right) \\ $$$$\int{lnxdx} \\ $$$$=\int\left(\mathrm{1}×{lnx}\right){dx} \\ $$$$={xlnx}−\int\left({x}×\frac{\mathrm{1}}{{x}}\right){dx} \\ $$$$={xlnx}−{x}+{c} \\ $$

Commented by Rio Michael last updated on 13/Nov/19

thanks

$${thanks} \\ $$

Answered by ajfour last updated on 13/Nov/19

(d/dx)(xln x)=ln x+1  ⇒ d(xln x)= (ln x)dx+dx  Integrating    ∫d(xln x)=∫(ln x)dx+∫dx    xln x+c = ∫(ln x)dx + x  ⇒  ∫ln x dx = xln x−x+c .

$$\frac{{d}}{{dx}}\left({x}\mathrm{ln}\:{x}\right)=\mathrm{ln}\:{x}+\mathrm{1} \\ $$$$\Rightarrow\:{d}\left({x}\mathrm{ln}\:{x}\right)=\:\left(\mathrm{ln}\:{x}\right){dx}+{dx} \\ $$$${Integrating} \\ $$$$\:\:\int{d}\left({x}\mathrm{ln}\:{x}\right)=\int\left(\mathrm{ln}\:{x}\right){dx}+\int{dx} \\ $$$$\:\:{x}\mathrm{ln}\:{x}+{c}\:=\:\int\left(\mathrm{ln}\:{x}\right){dx}\:+\:{x} \\ $$$$\Rightarrow\:\:\int\mathrm{ln}\:{x}\:{dx}\:=\:{x}\mathrm{ln}\:{x}−{x}+{c}\:. \\ $$$$ \\ $$

Commented by Rio Michael last updated on 13/Nov/19

i appreciate sir

$${i}\:{appreciate}\:{sir} \\ $$

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