Question Number 172608 by Mikenice last updated on 29/Jun/22
Answered by FelipeLz last updated on 03/Jul/22
![f(x) = x−sin(x) ∫((xcos(x)−sin(x))/([x−sin(x)]^2 ))dx ∫((xcos(x)−x+x−sin(x))/([x−sin(x)]^2 ))dx ∫(([x−sin(x)]−x[1−cos(x)])/([x−sin(x)]^2 ))dx ∫((f(x)−xf′(x))/([f(x)]^2 ))dx = (x/(f(x)))+c = (x/(x−sin(x)))+c](https://www.tinkutara.com/question/Q172900.png)
$${f}\left({x}\right)\:=\:{x}−\mathrm{sin}\left({x}\right) \\ $$$$\int\frac{{x}\mathrm{cos}\left({x}\right)−\mathrm{sin}\left({x}\right)}{\left[{x}−\mathrm{sin}\left({x}\right)\right]^{\mathrm{2}} }{dx} \\ $$$$\int\frac{{x}\mathrm{cos}\left({x}\right)−{x}+{x}−\mathrm{sin}\left({x}\right)}{\left[{x}−\mathrm{sin}\left({x}\right)\right]^{\mathrm{2}} }{dx} \\ $$$$\int\frac{\left[{x}−\mathrm{sin}\left({x}\right)\right]−{x}\left[\mathrm{1}−\mathrm{cos}\left({x}\right)\right]}{\left[{x}−\mathrm{sin}\left({x}\right)\right]^{\mathrm{2}} }{dx} \\ $$$$\int\frac{{f}\left({x}\right)−{xf}'\left({x}\right)}{\left[{f}\left({x}\right)\right]^{\mathrm{2}} }{dx}\:=\:\frac{{x}}{{f}\left({x}\right)}+{c}\:=\:\frac{{x}}{{x}−\mathrm{sin}\left({x}\right)}+{c} \\ $$