$$I=\lim_{n\rightarrow \infty}\int_0^{2\pi}{\frac{\cos nx}{x+1}}\text{d}x $$
the answer is 0. This is what I've done so far.
$$ \begin{split} \int_0^{2\pi}{\frac{\cos nx}{x+1}}\text{d}x&=\frac{1}{n}\int_0^{2\pi}{\frac{1}{x+1}}\text{d}\left( \sin nx \right)\\ &=\left. \frac{\sin nx}{n}\cdot \frac{1}{1+x} \right|_{0}^{2\pi}+\int_0^{2\pi}{\frac{\sin nx}{n}\cdot \frac{1}{\left( 1+x \right) ^2}}\text{d}x\\ &=\int_0^{2\pi}{\frac{\sin nx}{n}\cdot \frac{1}{\left( 1+x \right) ^2}}\text{d}x \end{split}$$
On
There are many ways to approach this.
A less common, but good general technique to know, is to apply Bonnet's second mean value theorem for integrals which states that if $f$ is integrable and $g$ is decreasing then there exists $\xi \in [a,b]$ such that
$$\int_a^b f(x) g(x) \, dx = g(a) \int_a^\xi f(x) \, dx.$$
Hence,
$$\left|\int_0^{2\pi}\frac{\cos nx}{x+1} \,dx \right| = \frac{1}{1+0}\left|\int_0^{\xi}\cos nx \,dx \right| = \frac{|\sin n\xi - \sin 0|}{n} \leqslant \frac{1}{n} $$
and we see that the limit is $0$.
Following your intuition, an inequality is useful at the point you got to, namely $|\sin(x)| \leq 1$ Using that, we have $$\left|\frac{1}{n}\int_0^{2\pi} \frac{\sin(nx)}{(1+x)^2}dx\right| \leq\frac{1}{n}\int_0^{2\pi}\frac{1}{(x+1)^2}dx=\frac{1}{n}\frac{-1}{x+1}\bigg|_0^{2\pi}=\frac{1}{n}\left(1-\frac{1}{2\pi+1}\right)<\frac{1}{n}$$ Now letting $n$ go to infinity we get $$\lim_{n \to \infty} |I| <\lim_{n \to \infty} \frac{1}{n} = 0$$ So $I=0$