$$\int_{0}^{\phi}{(1-x+x^2-x^3)^{2N-1}\over\left[1-(2N+1)x+(4N+1)x^2-(6N+1)x^3\right]^{-1}}\mathrm dx =5^{N}\phi\tag1$$
suppose $N\ge1$ and $\phi$ is the golden ratio.
How do we show that it does evaluate to $5^N\phi?$
Substitution doesn't seem to work for this integral. Unless we change these into a sum but how?
$$1-(2N+1)x+(4N+1)x^2-(6N+1)x^3$$ and
$$1-x+x^2-x^3$$
Let $f(x)=1-x+x^2-x^3$.
Then, $$1-(2N+1)x+(4N+1)x^2-(6N+1)x^3$$ can be written as $$\begin{align}&1-(2N+1)x+(4N+1)x^2-(6N+1)x^3\\\\&=1-x+x^2-x^3+2Nx(-1+2x-3x^2)\\\\&=f(x)+2Nxf'(x)\end{align}$$
Using this, we have $$\begin{align}\int_{0}^{\phi}f^{2N-1}(x)(f(x)+2Nxf'(x))\ \mathrm dx&=\int_{0}^{\phi}\left(f^{2N}(x)+x\left(f^{2N}(x)\right)'\right)\ \mathrm dx\\\\&=\int_{0}^{\phi}\left(xf^{2N}(x)\right)'\ \mathrm dx\\\\&=\left[xf^{2N}(x)\right]_{0}^{\phi}\\\\&=\phi(1-\phi+\phi^2-\phi^3)^{2N}\\\\&=5^N\phi\end{align}$$
The last equality comes from that $$1-\phi+\phi^2-\phi^3=(1-\phi)(1+\phi^2)=(1-\phi)(2+\phi)=2-\phi-(\phi+1)=-\sqrt 5$$ where we used that $\phi^2=1+\phi$.