A method to evaluate an integral such as $$I=\int_0^1 \frac{\ln(1+x)}{1+x^2}dx$$ is to make the substitution $x=\frac{1-u}{1+u}$ to receive $$I=\int_0^1 \frac{\ln2}{1+x^2} dx-I$$ which makes the integral trivial. I am trying to figure out the inverse problem. Given a differentiable bijection $\phi : [a,b] \to [a,b]$. I am trying to find an integral $$I=\int_a^b f(x) dx$$ that can be evaluated by the sustitution $x=\phi(u)$. After this substitution I want something of the form $$I=\int_a^b g(x)dx +cI$$ where $g$ has an elementary primitive. This gives the equation $$f\circ \phi \cdot \phi'=g+cf$$ which is equivalent to the condition that $$f\circ \phi \cdot \phi'-cf$$ has an elementary primitive. Now from Liouville's theorem we get $$f\circ \phi \cdot \phi'-cf= \sum_i c_i \frac{g_i'}{g_i}+h'$$ where $g_1,..., g_n $ and $h$ are in an elementary field containing $ f$ and $ \phi$
Looking back at my example up the top where $\phi(x)=\frac{1-x}{1+x}$, we get the condition that $$\frac{-2}{(1+x)^2}f\left( \frac{1-x}{1+x} \right ) -cf(x)= \sum_i c_i \frac{g_i'}{g_i}+h'$$ where $g_1,..., g_n $ and $h$ are in $C(x,\frac{1-x}{1+x},f)$
Which I have no idea how to attack except already knowing a solution. This question is more of an exploration, so any links to resources on the subject or ideas related to it are welcome.
Just to be clear my question is how I can create an integral that can be evaluated by a particular substitution. I have outlined my (incomplete) approach and given an example of the type of thing I am looking for.
What you have is a linear functional equation
$$ g(x) = \phi'(x) f(\phi(x)) - c f(x) \tag 1$$
In general functional equations are hard to solve, but there are some nice cases:
The function $\phi(x) = \frac{1-x}{1+x}$ is a nice example because it's an involution: $\phi(\phi(x)) = x$. Substituting $\phi(x)$ for $x$ in the functional equation gives
$$ g(\phi(x)) = \phi'(\phi(x)) f(x) - c f(\phi(x)) = \frac{f(x)}{\phi'(x)} - c f(\phi(x)) \tag 2$$
and you can solve (1) and (2) for $f(x)$: if $c^2 \ne 1$,
$$ f(x) = \frac{c g(x) + \phi'(x) g(\phi(x))}{1 - c^2} $$
However, in your example you want $c = -1$, which makes the system singular, and imposes a condition $g(x) = \phi'(x) g(\phi(x))$.