A peculiar Riemann Sum Im trying to express the following limit as a Riemann Sum
$$ \lim_{ n \to \infty} \sum_{i=1}^{n} \dfrac{ n^3 (2i-3) }{ (n^2- \sqrt{2n^4 + i^2n^2} + in )(2n^2-i^2)^{3/2} } $$
over the interval $[2,3]$
Clearly, we want $\dfrac{1}{n}$ out of the sum as to take $\Delta x$ in account. We can divide both sides by $\dfrac{1}{n^5}$ and we obtain the 
$$ \lim_{ n \to \infty} \dfrac{1}{n}\sum_{i=1}^{n} \dfrac{  \dfrac{(2i-3)}{n} }{ (n^2- \sqrt{2n^4 + i^2n^2} + in )(2n^2-i^2)^{3/2} } \cdot \frac{1}{\dfrac{1}{n^2 \cdot n^3}}$$
Which reduces to 
$$ \lim_{ n \to \infty} \dfrac{1}{n}\sum_{i=1}^{n} \dfrac{  \dfrac{(2i-3)}{n} }{ \left( 1+ \dfrac{i}{n} - \sqrt{ 2 + \dfrac{i^2}{n^2} } \right)\left(2- \left( \dfrac{i}{n} \right)^2 \right) } $$
Now, this is almost done as we want our height x-value to be $x = 2 + \dfrac{i}{n}$. However, here is where I get stuck and I ask for some help. The term $\dfrac{2i-3}{n}$ in the numerator is the only term that is problematic here. How would we handle this?

According to Riemann sum
$$\int_{a} ^{b} f(x) \, dx=\lim_{n\to\infty} \frac{b-a} {n} \sum_{i=1}^{n}f\left(a+i\cdot\frac {b-a} {n} \right)$$
we need to express the sum as a function of $\left(2+\frac{i}{n}\right)$ that is
$$\lim_{ n \to \infty} \frac{1}{n}\sum_{i=1}^{n} \frac{  2\left(2+\frac{i}{n}\right)-4+\frac 3 n }{ \left( \left(2+ \frac{i}{n}\right)-1 - \sqrt{ \left(2 + \frac{i}{n}\right)^2-4\left(2+\frac in\right)+6 } \right)\left(-2+4\left(2+\frac i n\right)- \left(2+ \frac{i}{n} \right)^2 \right) }=$$
$$=\int_2^3\frac{2x-4}{(x-1-\sqrt{x^2-4x+6})(-2+4x-x^2)}dx$$
which is an improper integral since $x-1-\sqrt{x^2-4x+6}=0$ for $x=\frac 5 2$.
Refer also to


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*Perfect understanding of Riemann Sums
