Definite integral $\int \frac{1}{x} y(x) dx$

It is not possible to compute $\int_a^b \frac{y(x)}{x}dx$ knowing only $y(a)$ and $y(b)$. Take the example of $a=0$, $b=1$. We can choose two different functions that have the same values at the endpoints, for example $y_1(x)=x$ and $y_2(x)=x^2$. Now if it is possible to determine $\int_0^1 \frac{y_1(x)}{x}dx$ and $\int_0^1 \frac{y_2(x)}{x}dx$ based solely on the values of $y_1$, $y_2$ at the endpoints, the two integrals must have the same value (since $y_1(0)=y_2(0)=0$ and $y_1(1)=y_2(1)=1$). But $\int_0^1 \frac{y_1(x)}{x}dx =1 \neq \frac{1}{2} =\int_0^1 \frac{y_2(x)}{x}dx$. In short, the value of the integral depends heavily on what happens inside the interval of integration.