I'm going to assume that what you want is an example of an integral such that the sum you wrote down is a Riemann sum for that integral. If that's not what you meant (very likely, I fear), please clarify your question.
Let's start with ${1\over n+1}+{1\over n+2}+\cdots+{1\over n+n}={1\over n}\left({n\over n+1}+{n\over n+2}+\cdots+{n\over n+n}\right)$ The right side can be written as ${1\over n}\left({1\over1+(1/n)}+{1\over1+(2/n)}+\cdots+{1\over1+(n/n)}\right)$ Now let $f(x)=1/(1+x)$. Then the sum is $(1/n)(f(1/n)+f(2/n)+\cdots+f(n/n))$ So you are evaluating the function $f$ at $n$ evenly-spaced points between zero and one, namely, at $1/n,2/n,\dots,n/n$, and then taking the average of all those function values by adding them up and dividing by $n$. That means that the sum you started with is a Riemann sum for the integral of that function $f$ between zero and one: $\int_0^1{1\over1+x}\,dx$