Prove that $\frac1{n+1} < \int_n^{n+1}\frac{1}{x}\mathrm dx$. This is probably a basic calculus question, but I have absolutely no idea where to go with this question. 

Prove:
  $$
\dfrac{1}{n+1}<\int_{n}^{n+1}{\dfrac{1}{x}dx}=\ln(n+1)-\ln(n)
$$

The antiderivatives were given as a hint I presume, but that seems basic enough and I'm not sure how it could be helpful.
Edit: I'm going to put the answer in here as use for anyone who comes along in the future and can't figure it out. I'm putting it in spoiler blocks because you should really try it yourself.
Proof:

 \begin{align} \dfrac{1}{n+1}<\dfrac{1}{x}<\dfrac{1}{n} \quad\textrm{for}\quad n<x<n+1 \end{align}
 \begin{align} \dfrac{1}{n+1}<\dfrac{1}{x} \implies \int_{n}^{n+1}{\dfrac{1}{n+1}dx}<\int_{n}^{n+1}{\dfrac{1}{x}dx} \end{align}
 \begin{align} \int_{n}^{n+1}{\dfrac{1}{n+1}} = {\dfrac{1}{n+1}x}\rvert{_{n}^{n+1}} = \dfrac{1}{n+1}(n+1-n)=\dfrac{1}{n+1}\end{align}
 \begin{align} \therefore\dfrac{1}{n+1}<\int_{n}^{n+1}{\dfrac{1}{x}dx} \end{align}

Q.E.D.

HINT:
For $x\in[n,n+1]$
$$\frac1{n+1}\le \frac1x\le \frac1n$$
and $\int_n^{n+1}(1)\,dx=1$.
