$\int_{0}^{1}{(1-x)(1-2x^{\phi})+\phi(x-x^{\phi})\over (1-x)^2}\cdot{\left(1-x^{\phi}\over 1-x\right)^{1\over \phi}}\mathrm dx=\phi^{\phi}$ How can we show that?

$$\int_{0}^{1}{(1-x)(1-2x^{\phi})+\phi(x-x^{\phi})\over (1-x)^2}\cdot{\left(1-x^{\phi}\over 1-x\right)^{1\over \phi}}\mathrm dx=\phi^{\phi}\tag1$$
  Where $\phi$;Golden ratio

This integral look too complicated. I try and make a $u=1-x$ still not simplified 
$$\int_{0}^{1}{u[1-2(1-u)^{\phi}]+\phi[1-u-(1-u)^{\phi}]\over u^2}\cdot\left[1-(1-u)^{\phi}\over u\right]^{1\over \phi}\mathrm du$$
$$\int_{0}^{1}{\phi-{u\over \phi}-(1-u)^{\phi}(2u+\phi)\over u^2}\cdot\left[1-(1-u)^{\phi}\over u\right]^{1\over \phi}\mathrm du$$
Simplifed $(1)$:
$$\int_{0}^{1}{1+{x\over \phi}+x^{\phi}(2x-\phi{\sqrt{5}})\over (1-x)^2}\cdot\left({1- x^\phi}\over 1-x\right)^{1\over \phi}\mathrm dx=\phi^{\phi} \tag2$$
I have no idea where to go from here.
 A: Using the relation $\phi - 1 = 1/\phi$, we find that
\begin{align*}
&\int_{0}^{1} \frac{(1-x)(1-2x^{\phi})+\phi(x-x^{\phi})}{(1-x)^2}\cdot\left(\frac{1-x^{\phi}}{1-x}\right)^{1/\phi} \, \mathrm{d}x \\
&= \int_{0}^{1} \left( 2 - \frac{\phi^2}{1-x^{\phi}} + \frac{\phi}{1-x} \right) \left(\frac{1-x^{\phi}}{1-x}\right)^{\phi} \, \mathrm{d}x \\
&= \bigg[ x \left(\frac{1-x^{\phi}}{1-x}\right)^{\phi} \bigg]_{0}^{1} \\
&= \phi^\phi.
\end{align*}
As a corollary, we have
$$ \int \frac{(1-x)(1-2x^{\phi})+\phi(x-x^{\phi})}{(1-x)^2}\cdot\left(\frac{1-x^{\phi}}{1-x}\right)^{1/\phi} \, \mathrm{d}x = x \left(\frac{1-x^{\phi}}{1-x}\right)^{\phi} + C. $$

Here is my line of reasoning that led to this solution:


*

*I tried to simplify the integrand so that it minimizes the amount of cancellation as well as mimics partial fraction decomposition.

*Now, the integrand in the second line looks similar to what we obtain when we apply the logarithmic differentiation
$$ \frac{d}{dx}\left(\frac{1-x^{\phi}}{1-x}\right)^{\phi} = \left( \frac{\phi}{1-x} - \frac{\phi^2 x^{\phi-1}}{1-x^\phi} \right) \left(\frac{1-x^{\phi}}{1-x}\right)^{\phi}. \tag{1} $$
Although this is not exactly the same as what we want, it hints that we might actually compute the antiderivative.

*Playing a little bit, we find that
$$ \frac{d}{dx}\frac{(1-x^{\phi})^{\phi}}{(1-x)^{\phi-1}} = \left( -2 -\frac{\phi^2 x^{\phi-1}}{1-x^{\phi}} + \frac{\phi^2}{1-x^{\phi}} \right) \left(\frac{1-x^{\phi}}{1-x}\right)^{\phi}. \tag{2}$$
Bingo! $\text{(1)} - \text{(2)}$ gives exactly what we want and we are done.
