Question

интеграл (1-cosx)/((x-sinx)^2)

Answer 1

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Answer 2

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