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?

You're so close to having the answer!
You've calculated expressions for $F(ax)$ up to constants of integration in two ways, so the two expressions must differ by a constant (of integration). That is, there are constants $C_1, C_2$ such that
$$bF(x) + C_1 = \frac{1}{a}F(ax) + C_2$$
Now just multiply through by $a$ and rename your constants appropriately.
