Prove that if $\int_0^1 f(yx) dx = g(y)$, so $f(x)=g(x)+xg'(x)$ for any $x$. I'm trying to solve this question:

$f: \mathbb{R} \to \mathbb{R}$ is a continuous function and $g: \mathbb{R} \to \mathbb{R}$ is of $C^1$ class, such that $g(y) = \int_0^1 f(yx) dx$. Prove that $f(x)=g(x)+xg'(x)$ for all $x \in \mathbb{R}$.

But my progress until now was really slow. I've tried to use the Leibniz integral rule and the first mean value theorem for definite integrals, but I had no success. Can somebody help me?

If you change variables in the integral ($u = xy$), you get
$$
\int_0^1 f(yx) dx = \frac 1y \int_0^yf(u) du,
$$
so,
$$\int_0^y f(u) dy = yg(y) \Rightarrow f(y)=(yg(y))' = g(y) + yg'(y).$$
Note: This method requires that $y\ne 0$, but in that case the equality can be checked directly.
