Let \int_1^x f(x) dx = g(x)
Then the given equation may be written as
g(xy) = yg(x)+xg(y)
or \frac{g(xy)}{xy} = \frac{g(x)}{x}+\frac{g(y)}{y}
Hence h(x) = \frac{g(x)}{x} satisfies the well known functional equation h(xy) = h(x)+h(y) for which the solution is h(x) = c ln x.
Hence \int_1^x f(x) \ \dx = cx \ln x \Rightarrow f(x) = c (\ln x+1)