Computing $f(x)$ if $\int_1^{xy} f(t) dt = y \int_1^x f(t) dt + x\int_1^y f(t) dt$ and $f(1) = 3$ This is a problem from Apostol, Calculus, Volume I, Chapter 6.9 (p. 238), that I was hoping someone could help with:

A function $f$, continuous on the positive real axis, has the property that $$\int_1^{xy} f(t) dt = y \int_1^x f(t) dt + x \int_1^y f(t) dt$$
  for all $x > 0$ and all $y > 0$.  If $f(1) = 3$, compute $f(x)$ for each $x > 0$.

My initial thought was to use properties of the integral to write:
$$\begin{align*} 
\int_1^{xy} f(t) dt & = \int_1^x f(t) dt + \int_x^{xy} f(t) dt \\
& = \int_1^x f(t) dt + x \int_1^y f(xt) dt
\end{align*}$$
and
$$\begin{align*}
\int_1^{xy} f(t) dt & = \int_1^y f(t) dt + \int_y^{xy} f(t) dt \\
& = \int_1^y f(t) dt + y \int_1^x f(yt) dt
\end{align*}$$
Then each of these is also $=y \int_1^x f(t)dt + x \int_1^ f(t)dt$ from the given equation of the problem.  I cannot seem to pull an equation for $f(x)$ out of this though (I might be missing some very simple manipulation...).
I also thought in terms of letting $A(x) = \int_1^x f(t)dt$, so the problem is giving us a functional equation:
$$ A(xy) = yA(x) + xA(y)$$
This has obvious similarities to the functional equation of the logarithm.
This problem is in the section of the text with the definition of the logarithm as an integral.  I'm self-studying, so this isn't homework.  Hints or full solutions are equally welcome.  Thanks for the help.
 A: Fix $x$ and differentiate both sides of your equation with respect to $y$ to obtain:
$$\tag{1}
x f(xy)= \int_1^xf(t)\,dt +xf(y).
$$
Set $y=1$ in equation $(1)$ to obtain
$$\tag{2}
xf(x)=3x+\int_1^x f(t)\,dt.
$$
From $(2)$, we see that $f$ is differentiable.
Unfixing $x$ now, differentiate both sides of $(2)$ with respect to $x$ to obtain
$$
 f(x)+xf'(x)= 3+f(x).
$$
So 
$$\tag{3}
xf'(x)=3; \ \ f(1) =3.
$$
The solution of the initial value problem $(3)$ is $f(x)= 3\ln x +3$.
A: $$A(xy) = yA(x) + xA(y)$$
Can be rewritten as 
$$\frac{A(xy)}{xy}=\frac{A(x)}{x}+\frac{A(y)}{y} \,.$$
Let $B(x)=\frac{A(x)}{x}$ .
Then $B(xy)=B(x)+B(y)$. Which means that
$$B(e^{t+s})=B(e^{t}e^{s})= B(e^{t})+B(e^{s})\,.$$
Setting $C(t)=B(e^{t})$ you get that $C$ satisfies the Cauchy Equation 
$$C(t+s)=C(t)+C(s) \,.$$
P.S. The equation of $B(x)$ is exactly the equation of the logarithm, you can actually stop there and figure which log it is...
A: From OP's work. A(xy)=xA(y)+yA(x) \implies A(z)=Cz \ln z$. Nex, $A(x)=\int_{1}^{x} f(t) dt \implies Cx\ln x=\int_{1}^{x} f(t) dt$$ differentiating using Lebnitz, we get
$f(x)=C(1+\ln x)$ and  $f(1)=3$ gives $C=3$. So finally $f(x)=3(1+\ln x).$
