How would I solve integrals of derivatives? 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.
$$I=\int\limits_a^b\int\frac{\mathrm df(t)}{\mathrm dt}\mathrm df(t)\mathrm dt$$


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 $$\iint f'(t)^2\mathrm dt^2.$$
$$I=\int\limits_a^b\int\frac{\mathrm df(t)}{\mathrm dt}\mathrm df(t)\mathrm dt$$


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 $$\iint f'(t)^2\mathrm dt^2.$$