$\lim_{n \to \infty}(\frac{1}{\sqrt {n^2+1}}+\frac{1}{\sqrt {n^2+2}}+\cdots+\frac{1}{\sqrt {n^2+n}})$ If we have $(n^2+a^2)$ instead of $(n^2+a)$ in the denominator, we just evaluate ${\int_{0}^{1}{\frac1{\sqrt{1+x^2}}dx}}$. But how do I express this above form as a definite integral?

You want to calculate
$$X = \lim_{n\to\infty} \left( \sum_{k=1}^n \frac{1}{\sqrt{n^2+k}} \right)$$
You can estimate it from both sides:
$$\lim_{n\to\infty} \left( \sum_{k=1}^n \frac{1}{\sqrt{n^2}} \right) \leq X \leq 
  \lim_{n\to\infty} \left( \sum_{k=1}^n \frac{1}{\sqrt{n^2+n}} \right)$$
This can be simplified to
$$1 = \lim_{n\to\infty} \left( \frac{n}{\sqrt{n^2}} \right) \leq X \leq 
  \lim_{n\to\infty} \left( \frac{n}{\sqrt{n^2+n}} \right) = 1$$
so, the solution is $X=1$.
