I'm stuck with this exercise:
Let $f\colon (0,+\infty) \to \mathbb{R}$ such that $f(x)=\int_{x}^{x+\sin{x}}\frac{dt}{\log(1+t)}.$ Prove that $\lim _{x \to +\infty} f(x)=0.$
All I have found is that $\int_{x}^{x+\sin{x}}\frac{dt}{t}\leq f(x) \; \forall x\geq0$ and $\lim _{x \to \infty} \int_{x}^{x+\sin{x}}\frac{dt}{t}=0$, although I can't find an upper bound for $f(x)$.
Using the fact that for all $x > 0$, $$ |f(x)| \leq \int_{x-1}^{x+1} \frac{dt}{\log(1+t)} \leq \frac{1}{\log(x)}\int_{x-1}^{x+1} dt $$ because we integrate in $t \geq x$ and the logarithm is non-decreasing. Hence, $$ \lim_{x\to \infty} f(x) \leq \lim_{x\to \infty} \frac{2}{\log(x)} = 0. $$