Solving integrals of values like $\int f \, \mathrm{d} f $ is simple enough but how would I solve integrals of derivatives?
For example consider the integral:
$$ \Delta F = \int^b_a \int \frac{\mathrm{d} f}{\mathrm{d} t} \, \mathrm{d} f \, \mathrm{d} t $$
I guess this can be broken up into two parts.
$$ \frac{\mathrm{d} u}{\mathrm{d} t} = \frac{\mathrm{d} f}{\mathrm{d} t} \frac{\mathrm{d} f}{\mathrm{d} t} $$
and
$$ \Delta F = \int^b_a u \, \mathrm{d} t $$
I'm interested in this because its functional derivative is (I think):
$$ \frac{\delta F}{\delta f} = \frac{\partial u}{\partial f} - \frac{\mathrm{d}}{\mathrm{d} t}\frac{\partial u}{\partial f'} $$
$$ \frac{\delta F}{\delta f} = 0 - 2\frac{\mathrm{d} f}{\mathrm{d} t}$$
but that doesn't seem right to me and it should be something more like $\frac{\mathrm{d} f}{\mathrm{d} t}$.
Using a polynomial series I can represent the function as:
$$ \left[ Dt + \sum^{\infty}_{k=1} \sum^{\infty}_{l=1} \frac{c_k c_l k l}{(k + l -1)(k + l)} t^{k + l} \right]^b_a $$
but I don't know if I can deduce any useful properties from that.
Notice that $$\mathrm df(t)=\frac{\mathrm df(t)}{\mathrm dt}\mathrm dt,$$ so $$I=\int\limits_a^b\int\frac{\mathrm df(t)}{\mathrm dt}\frac{\mathrm df(t)}{\mathrm dt}\mathrm dt\mathrm dt=\int\limits_a^b\int\left(\frac{\mathrm df(t)}{\mathrm dt}\right)^2\mathrm dt\mathrm dt=\int\limits_a^b\int f'\left(t\right)^2\mathrm dt\mathrm dt.$$
There most probably is no simpler form for the second antiderivative of the square of the first derivative than just $$\iint f'(t)^2\mathrm dt^2.$$