As the function in the integral is continuous is \(\displaystyle [2,7]\) we know the integral exists and thus we can choose the partition of the interval as we want. We choose

to subdivide \(\displaystyle [2,7] \) in n subintervals of equal length \(\displaystyle \frac{7-2}{n}=\frac{5}{n}\) , and we evaluate the function at right end of each subinterval:

\(\displaystyle \lim_{n\to\infty}\sum^n_{i=i}\frac{5}{n}\left[2+\frac{5i}{n}-3\ln\left(2+\frac{5i}{n}\right)\right]\) ...this looks similar to what you got but I can't be sure since it isn't clear the way you wrote it.