Flip side of Feynman's trick for Integration If I differentiate the integral:
$$\int_{-a+2}^{a-2} \ (a-x) \, da$$
then I get 4 - 2 a.
1) Is it possible to get back to integral in the form $ \int_{-a+2}^{a-2} \ (a-x) \, da$?
The application would be to find a way to use the 'flip side of Feynman's trick' described on page 90 and 91. The author Paul J. Nahin of Inside Interesting Integrals appears to find the integral that when integrated again (double integration) leads to the solution. So I thought if one differentiated the original definite integral one could find what the integral should be.  Otherwise the integral seems to have to be guessed.
To illustrate what I'm getting at, so he finds:
$$\int_{0}^{1}   \frac{x^a-1}{\ln(x)}\,dx\,=ln(a+1)\, a > 0, \, a = 0$$
by using:
$$\int_{0}^{1} \ x^y  dy\, = \frac{x^a-1}{\ln(x)}\,$$
So I thought could one differentiate
$$\int_{0}^{1} \frac{x^a+1}{\ln(x)} \, dx\, a > 0, \, a = 0$$
to get:
$$\int_{0}^{1} \ x^y  dy\,$$
because without him saying so I cannot see how one could guess this integral.
Hence the question 1.

$x$ is a bound (also called dummy) variable. You can use any other variable except $a$
$$
\int_{-a+2}^{a-2}\ (a-x)\,dx =
\int_{-a+2}^{a-2}\ (a-z)\,dz = \cdots
$$
In any case, your integral only depends of $a$.
Edit:
$$
\frac{d}{da}\int_0^1\frac{x^a - 1}{\ln(x)}\,dx\,=
\int_0^1 x^a\,dx
$$
Are you mixing Leibniz rule (the kernel of Feynman's trick) with the Fundamental Theorem of Calculus?
