Prove or disprove convergence of $\int_0^1 \frac{dx}{x-sinx}$ Prove or disprove convergence of $\int_0^1\frac{dx}{x-sinx}$
Because $x=0$ is a problem here as $x\to0$ $sinx=x-\frac{x^3}{6}+o(x^3)$
So $\frac{1}{x-sinx}\approx\frac{6}{x^3+o(x^3)}$
Now how here get rid of $o(x^3)$ to use ratio test?
Or maybe there is an other easy way to prove divergence of given integral

You don't need to get rid of the $o(x^3)$ term. Your estimate is correct and suggests that the integral behaves roughly like $\int_0^1 \frac{1}{x^3} dx$ which diverges. To prove this rigorously we need to use the comparison test:
$$\lim_{x\to 0}\frac{\frac{1}{x-\sin x}}{\frac{1}{x^3}}=\lim_{x\to 0}\frac{x^3}{x-\sin x}=\lim_{x\to 0} \frac{3x^2}{1-\cos x}=\lim_{x\to 0}\frac{6x}{\sin x}=6$$
where L'Hopital's rule was used a few times. This justifies the intuition and proves that your integral diverges, since $\int_0^1 \frac{1}{x^3} dx$ diverges.
