Derivating an integral I have a question about, how can i derivate an expression of the form $$
F\left( u \right) = \int\nolimits_{C_1 }^{C_2 } {f\left( {ux} \right)\mathrm{d}x} 
$$
where $
C_1 ,C_2 
$
are constants.
I have no idea :/
 A: This operation is called differentiating under the integral sign. In the general case, 
$$
F(u) = \int_{a(u)}^{b(u)} g(u,x) dx,
$$
where the limits are functions of the parameter $u$, we have
$$
F'(u) = f(u, b(u)) b'(u) - f(u, a(u)) a'(u) + \int_{a(u)}^{b(u)} \frac{\partial}{\partial u} g(u,x) dx.
$$
In your case, the limits are constants, and $g(u,x)$ has the special form $f(ux)$. So the first two terms drop out, giving:
$$
F'(u) = \int_{C_1}^{C_2} \frac{\partial}{\partial u} f(ux) dx = \int_{C_1}^{C_2} f'(ux) x dx,
$$
thanks to the chain rule. 
In general, one cannot "simplify" such answers any further, but in your case you can. Noting that 
$$
F'(u) = \int_{C_1}^{C_2} \frac{x}{u} \frac{\partial}{\partial x} f(ux) dx,
$$
and integrating by parts, we get:
$$
F'(u) 
= \left. \frac{x}{u} f(ux) \right|_{C_1}^{C_2} - \int_{C_1}^{C_2} \left(\frac{\partial}{\partial x} \frac{x}{u} \right) \cdot f(ux) dx 
= \ldots
$$
I will leave it to you to complete the answer. You should be able to express the final integral in terms of $F(u)$ itself.
Edit: As @joriki notes in a comment, the "simplification" step is valid only if $u \neq 0$.
