I know that there are already a couple of questions regarding this topic here on MathSE but none could give me a satisfying answer so far...
I am dealing with the integral $$\int \Big(f(x) + o(h^2x^2)\Big)\mathrm dx,$$ where $o$ is understood with regard to $h\rightarrow 0$. Apparently, the result is $$o(h^2) + \int f(x),$$ assuming that $\int x^2f(x) = const$. So it seems like little $o$ and integral were interchanged . But how can that be justified?
Remember $f(h) = o(g(h))$ if:
$\begin{align*} \lim_{h \to 0} \frac{f(h)}{g(h)} &= 0 \end{align*}$
By the definition of the limit (assuming positive functions!) this means that for any $\epsilon > 0$ there is a $\delta$ such that for $\lvert h \rvert \le \delta$:
$\begin{align*} \left\lvert \frac{f(x)}{g(x)} \right\rvert &\le \epsilon \\ f(h) &\le \epsilon g(h) \end{align*}$
In your case:
$\begin{align*} u(x) &= o(x^2 h^2) \\ u(x) &\le \epsilon x^2 h^2 \\ \int_{x_0}^{x_1} u(x) \, d x &\le \epsilon h^2 \int_{x_0}^{x_1} x^2 \, d x \\ &= \epsilon h^2 \frac{x_1^3 - x_0^3}{3} \\ &= o(h^2) \end{align*}$