Can we show that the result of an integral of a function of polynomial order is always one degree higher? Is it true that if $$ \int f(x)\,\mathrm dx = g(x) $$
for $f\in\mathcal O(x^n)$, then $g$ must be in $\mathcal O(x^{n+1})$ where $\mathcal O$ denotes big-O notation and $ n \ge 1 $ is an integer?
If it is true, how can we prove it?
If it is false, can we have a counterexample as the answer?

(I've decided to delete my old answer since I had made a step that was false. I decided that the proof was unsalvageable and wrote this proof instead)
For some $n>-1$ and $f(x)\in O\left(x^n\right)$, suppose that $$g(x)=\int f(x)dx\notin O\left(x^{n+1}\right)$$
Now this implies that there does not exist real $k$ such that, provided $x>x_0$ for some real $x_0$, $$g(x)\leq kx^{n+1}$$
Or alternatively, for every real $k$ and $x>x_0$ we have $$g(x)\leq kx^{n+1}$$
Taking the derivative:
$$\frac d{dx}g(x)>\frac d{dx}kx^{n+1},x>x_0$$
$$f(x)>k(n+1)x^n,x>x_0$$
However, since $f(x)\in O\left(x^n\right)$, this implies that there exists $M$ and real $x_1$ such that $$f(x)\leq Mx^n,x>x_1$$
Let $k=\frac M{n+1}$. The equation we obtained by taking the derivative becomes $$f(x)>Mx^n,x>x_0$$
But when $x>x_0$ and $x>x_1$ then we have
$$Mx^n\geq f(x)>Mx^n$$
Clearly a contradiction.
$\therefore$ $g(x)\in O\left(x^{n+1}\right)$
