Question

интеграл (1-cosx)/((x-sinx)^2)

Answer 1

.....................

Answer 2

$\int \frac{(1-cosx)dx}{(x-sinx)^2}=[\, t=x-sinx\; ,\; dt=(x-sinx)'dx=(1-cosx)dx\, ]=\\\\=\int \frac{dt}{t^2}=\int t^{-2}\, dt=\frac{t^{-1}}{-1}+C=-\frac{1}{x-sinx}+C$