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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 :/

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    @August, The new example is **not** of the form you gave in your question. And the differentiation seems ok. Note that you are (partial) differentiating w.r.t. $a$, not $x$. So the $x$ is$a$constant. Also differentiating $\arctan (ax)$ w.r.t. $a$ gives a spare $x$, thanks to chain rule. These two $x$'s cancel nicely.2011-09-04

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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 $u \neq 0$.