Getting different answer when evaluating an integral from a released exam. I am working on problem number 3a on this exam.

Suppose $f$ is a continuous function on $(0,+\infty)$, and $a$ and $b$ are positive real numbers such that $f(ax)=bf(x)$ for all $x>0$.  Show that if $F$ is any antiderivative of $f$, there will exist a constant $k$ such that $F(ax)=abF(x)+k$ for all $x>0$.

My thought process was:
$$F(ax)=\int f(ax)\, dx=\int bf(x)\,dx=b\int f(x)\,dx=bF(x)+C$$
Change $C$ to $k$ to adapt to the question:
$$ = bF(x) + k $$
Also tried doing U-substitution:
$$F(ax)=\int f(u)\,du$$
where $u=ax$ and $du=a\,dx$
which gives:
$$\frac{1}{a}F(ax) + C$$
but my way of doing it doesn't have the $a$ in the end result. I feel like I'm probably forgetting some simple rule but I can't remember it. Where did I go wrong?

Here's another way to think about it, which is probably equivalent to your argument (completed by Clive):
We want to prove that there exists a constant $k$ such that $F(ax) = abF(x) + k$. Let $L(x) = F(ax)$ and $R(x) = abF(x)$ ($L$ and $R$ for left and right, respectively). Then $L'(x) = R'(x) = abf(x)$. Since $L$ and $R$ have the same derivative, they must differ by a constant.
