Answer
Diverges
Work Step by Step
Given $$ \int_{0}^{1}\frac{dt}{t-\sin t}$$
Let $$f(t) =\frac{1}{t-\sin t} ,\ \ \ g(t) = \frac{1}{t^3}$$
Then
\begin{align*}
\lim _{t \rightarrow 0} \frac{f(t)}{g(t)}&=\lim _{t \rightarrow 0} \frac{t^{3}}{t-\sin t}\\
&=\lim _{t \rightarrow 0} \frac{3 t^{2}}{1-\cos t}\\
&=\lim _{t \rightarrow 0} \sin t\\
&=\lim _{t \rightarrow 0} \frac{6}{\cos t}\\
&=6
\end{align*}
and
\begin{align*}
\int_{0}^{1} \frac{d t}{t^{3}}&=\lim _{b \rightarrow 0^{+}}\left[-\frac{1}{2 t^{2}}\right]_{b}^{1}\\
&=\lim _{b \rightarrow 0^{+}}\left[-\frac{1}{2}-\left(-\frac{1}{2 b^{2}}\right)\right]=+\infty
\end{align*}
Hence $$ \int_{0}^{1}\frac{dt}{t-\sin t}$$
also diverges