I want to solve the following integral (which has to depend on $u$) $$I(u)=\int_{u_1}^u \frac{f(x)}{x(1-x)}dx,~~~~~~u_1=const$$ I would like to know if I can apply the mean value theorem like this: $$I(u)=\int_{u_1}^u \frac{f(x)}{x(1-x)}dx=f(\xi)\int_{u_1}^u\frac{1}{x(1-x)}dx$$
If this is right, how can I compute $f(\xi)$? If I do: $$f(\xi)=\frac{\int_{u_1}^{u_2}\frac{f(x)}{x(1-x)}dx}{\int_{u_1}^{u_2}\frac{1}{x(1-x)}dx}$$ I get (obviously) different values depending on $u_2$. I would like a fix expression (or value) for $f(\xi)$ to be able to do my integral.
Thanks in advance for your ideas.