integration - Help on simplify $int_0^1frac{(xphi)^n-(-1)^n}{phi^{2n}-(-1)^n}cdot(x^n-phi^n)dx$

$\phi=\frac{1+\sqrt5}{2}$

Fibonacci numbers

$F_1=1$, $F_2=1$ ; $F_{n+2}=F_{n+1}+F_{n}$

A first few values for $n=1,2,3,4,...$ of $F_n$ are $1,2,3,5,...$

Lucas numbers

$L_1=1$, $L_2=3$ ; $L_{n+2}=L_{n+1}+L_n$

A first few values for $n=1,2,3,4,...$ of $L_n$ are $1,3,4,7...$

Show that,

$$\int_0^1\frac{(x\phi)^n-(-1)^n}{\phi^{2n}-(-1)^n}\cdot(x^n-\phi^n)dx=\left[\frac{1}{2n+1}-\frac{L_n}{n+1}+(-1)^n\right]\cdot\frac{1}{\sqrt5F_n}$$

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$$\int_0^1\frac{(x\phi)^n-(-1)^n}{\phi^{2n}-(-1)^n}\cdot(x^n-\phi^n)dx=\frac{1}{\phi^{2n}-(-1)^n}\left[\frac{\phi^n}{2n+1}-\frac{\phi^{2n}}{n+1}-\frac{(-1)^n}{n+1}+(-1)^n\phi^n\right]$$

I am stuck, can't simplify to get the proposed result. Can anybody help me please?

Well we know that $\phi^n={\phi}F_n+F_{n-1}$. This didn't help, it makes more complicated. There must be an easy way for this, but can't figured out yet.

Binet's formula $F_n=\frac{\phi^n-(-\phi)^{-n}}{\sqrt5}$ and

$L_n=\phi^n+(-\phi)^{-n}$

Answer

Let's write $\psi = \dfrac{-1}{\phi} = \dfrac{1-\sqrt{5}}{2}$.

The trick is to multiply numerator and denominator of $\dfrac{(x\phi)^n - (-1)^n}{\phi^{2n} - (-1)^n}$ with $\phi^{-n}$ to get

$$\frac{(x\phi)^n - (-1)^n}{\phi^{2n} - (-1)^n} = \frac{x^n - \psi^n}{\phi^n - \psi^n} = \frac{x^n - \psi^n}{\sqrt{5}\,F_n}.$$

Now we can multiply the numerator with the remaining factor:

$$(x^n - \psi^n)(x^n - \phi^n) = x^{2n} - x^n\cdot (\phi^n + \psi^n) + (\phi\cdot \psi)^n = x^{2n} - L_n x^n + (-1)^n,$$

and the rest is straightforward.