How to compute infinite limits of summations (Riemann sums) when there are i's in the denominator? I'm really lost here. I'm trying to use a right Riemann sum to compute:
$$\int_0^1{\frac{x}{x^4+2x^2+1}dx}$$
Eventually I get here:
$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\frac{n^3i}{n^4+2n^2i^2+i^4}$$
Using $\Delta{x}=1/n$ and $x_i=a+i\Delta{x}=0+i/n=i/n$
But now I'm stuck since the only formulas I know to manipulate the summation and get rid of the i's are like $\sum_{i=1}^ni^2$. But since I have i's in the denominator, how can I proceed? Thanks!
I hope you are OK with the substitution $x^2+1=u$ to get the integral as $$\int_{1}^{2}\frac{du}{2u^2}$$ A standard technique now is to use a Riemann sum where the partition points are in geometric progression (in contrast to the more common arithmetic progression).
Let $x_i=2^{i/n}$ be the points of partition and we form the Riemann sum $$\sum_{i=1}^{n}f(x_i)(x_i-x_{i-1})$$ where $f(x) =1/(2x^2)$. The computation of the limit is not difficult and should give you the answer $1/4$.

If on the other hand you wish to use the partition points in arithmetic progression as $x_i=1+(i/n)$ then you need some approximation technique for the Riemann sum. Thus we have $$f(x_i) \Delta x_i=\frac{1}{2n(1+i/n)^2}$$ and $$\frac{1}{2}\left(\frac {1}{1+i/n}-\frac{1}{1+(i+1)/n}\right) \leq f(x_i)\Delta x_i\leq \frac{1}{2}\left(\frac{1}{1+(i-1)/n}-\frac{1}{1+i/n}\right)$$ Adding these inequalities for $i=1,2,\dots, n$ we get $$\frac{1}{2}\left(\frac{1}{1+1/n}-\frac{1}{1+(n+1)/n}\right)\leq \sum_{i=1}^{n}f(x_i)\Delta x_i\leq \frac {1}{2}\left(1-\frac{1}{2}\right)$$ By squeeze theorem the limit of Riemann sum as $n\to \infty $ is $1/4$.
The approximation used above can be applied to the original integral also (without the substitution $u=1+x^2$) but it is slightly more complicated to deal with.
As per @Aryadeva:
$$I=\int_{1}^{2} \frac{du}{2u^2}=\frac{1}{2}\lim_{n \to \infty} \sum_{k=n}^{2n} \frac{n^2}{k^2} \frac{1}{n}= \lim_{n \to \infty} \frac{n}{2} \sum_{k=n}^{2n} \frac{1}{k^2}=\lim_{n \to \infty}\frac{n}{2}[H^{(2)}_{2n}-H^{(2)}_n]=\lim_{n \to \infty}\frac{n}{2}[\psi^1(n)-\psi^1(2n+1)]~~~~~~(1)$$
Here $H^{(2)}_n$ are harmonic numbers and $\psi^m(x)$ are called polygamma functions.
Using the bounds on $\psi^m(z)$ as
$$ \frac{(m-1)!}{x^m}+\frac{m!}{2x^{m+1}} \le \psi^m(x) \le \frac{(m-1)!}{x^m}+\frac{m!}{x^{m+1}}~~~~(2) $$
we get $$ 1+\frac{1}{2n} \le n~\psi^1(n) \le  1+\frac{1}{n}$$
So by sandwich theorem (squeez law) $$\implies \lim_{n \to \infty} n~ \psi^1(n)=1~~~~(3)$$
and
$$ \frac{n}{2n+1} +\frac{n}{2(2n+1)^2}\le  n \psi^1(2n+1) \le \frac{n}{2n+1}+\frac{n}{(2n+1)^2}$$
Again $$\lim_{n \to \infty} n \psi^1(2n+1) =\frac{1}{2}~~~~~~(4)$$
Fimally using (3) and (4) in (1), we get
$$I=\frac{1}{4}.$$
For polygamma functions see
https://en.wikipedia.org/wiki/Polygamma_function