If $f(17) = 17$, calculate $A(x) = \int_{1}^{x} f(t)dt$ for $x>0$.

I guess you wanted to write $A(x) := \int_{1}^{x} f(t)dt$ instead of $A(x) = \int_{2}^{x} f(t)dt$. Then, as you had seen, we get the equation $A(xy)=A(x)+A(y)$, and so $$ \begin{align*} A(xy)=A(x)+A(y)&\implies \partial_y A(xy)=A'(xy)x=A'(y)\\ &\implies f(xy)x=f(y)\\ &\implies f(17x)=\frac{17}x\\ &\implies f(s)=\frac{(17)^2}s,\quad s:=17x\\ &\implies \int_1^x f(s)\,\mathrm d s=(17)^2\int_1^x\frac1s\,\mathrm d s\\ &\implies A(x)=(17)^2\ln(x) \end{align*} $$

We can differentiate the original identity with respect to $y$ to obtain: $$xf(xy) = \frac{d}{dy}\int_{x}^{xy}f(t)\,dt = \frac{d}{dy}\int_{x'}^{x'y}f(t)\,dt = x'f(x'y)$$ for all $x,x',y > 0$. Plugging in $x' = 17$ and $y = 1$ gives $$xf(x) = 17f(17) = 17^2 = 289$$ or $$f(x) = \frac{289}{x}$$ for all $x > 0$.

Now for all $x > 0$ we have

$$A(x) = \int_1^x f(t)\,dt = \int_1^x\frac{289}{t}\, = 289\ln x$$