Calculate $ \int_0^1{\frac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx$ 
Given a function $f(x)$ that is differentiable on $\left[0; 1\right]$
  satsifies: $$ f(1) = 1 $$ $$ f(x)f(1-x) = e^{x^2 - x} $$ 
  Calculate:  $$ \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx $$

Attempt number 1:
Using integration by parts, we have:
\begin{align}
\int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx &= \left(2x^3 - 3x^2\right)\Big|_0^1 - \int_0^1{\dfrac{\left(6x^2 - 6x\right)f(x) - \left(2x^3 - 3x^2\right)f'(x)}{\left[f(x)\right]^2}}f(x)\,dx\\
&= \left(2x^3 - 3x^2\right)\Big|_0^1 - \int_0^1{\left(6x^2 - 6x\right)}\,dx + \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx
\end{align}
This gives me equation $0 = 0$, in which I can't do anything.
Attempt number 2:
Express $f(x)$ in terms of $f(1-x)$:
$$ f(x) = \dfrac{e^{x^2-x}}{f(1-x)} $$
This implies that:
$$ f'(x) = \dfrac{2xf(1-x)e^{x^2-x} - f'(1-x)e^{x^2-x}}{\left[f(1-x)\right]^2} $$
Subtitute in, we have:
\begin{align}
\int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx &= \int_0^1{\dfrac{\left(2x^3-3x^2\right)\left(2xf(1-x) - f'(1-x)\right)}{f(1 - x)}}\,dx\\
&= \int_0^1{2x\left(2x^3-3x^2\right)}\,dx - \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(1-x)}{f(1 - x)}}\,dx\\
&= -\dfrac{7}{10} - \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(1-x)}{f(1 - x)}}\,dx
\end{align}
Then, I tried to turn $1 - x$ into $x$ in the last integral but failed to come up with anything useful.
I would like to know whether there is another way to solve this problem or how my second attempt could have been done.
Thanks in advance
 A: Denoting $ I=\int\limits_{0}^{1}{x\left(1-x\right)\ln{\left(f\left(x\right)\right)}\,\mathrm{d}x} $, we have :
\begin{aligned} \int_{0}^{1}{\frac{\left(2x^{3}-3x^{2}\right)f'\left(x\right)}{f\left(x\right)}\,\mathrm{d}x}&=\left[\left(2x^{3}-3x^{2}\right)\ln{\left(f\left(x\right)\right)}\right]_{0}^{1}-6\int_{0}^{1}{\left(x^{2}-x\right)\ln{\left(f\left(x\right)\right)}\,\mathrm{d}x}\\ &=6\int_{0}^{1}{x\left(1-x\right)\ln{\left(f\left(x\right)\right)}\,\mathrm{d}x}\\ \int_{0}^{1}{\frac{\left(2x^{3}-3x^{2}\right)f'\left(x\right)}{f\left(x\right)}\,\mathrm{d}x}&=6I \end{aligned}
Let's calculate $ I $, making the substitution $ y=1-x $, we get : $$ I=\int_{0}^{1}{\left(1-y\right)y\ln{\left(f\left(1-y\right)\right)}\,\mathrm{d}y} $$
Meaning : \begin{aligned} 2I&=\int_{0}^{1}{x\left(1-x\right)\ln{\left(f\left(x\right)\right)}\,\mathrm{d}x}+\int_{0}^{1}{\left(1-y\right)y\ln{\left(f\left(1-y\right)\right)}\,\mathrm{d}y}\\ &=\int_{0}^{1}{x\left(1-x\right)\ln{\left(f\left(x\right)f\left(1-x\right)\right)}\,\mathrm{d}x}\\ 2I&=-\int_{0}^{1}{x^{2}\left(1-x\right)^{2}\,\mathrm{d}x}=-\frac{1}{30} \end{aligned}
Thus, $$ \int_{0}^{1}{\frac{\left(2x^{3}-3x^{2}\right)f'\left(x\right)}{f\left(x\right)}\,\mathrm{d}x}=-\frac{1}{10} $$
A: From the given functional equation, we obtain that $f(0)=f(1)=1.$
Now setting $I=\int_0^1 \frac{(2x^3-3x^2)f'(x)}{f(x)}\mathrm dx$ and integrating by parts gives $$I=(2x^3-3x^2)\int \mathrm d\left(\log f(x)\right)-\int(6x^2-6x)\log f(x)\mathrm dx.$$ Substituting the limits gives that $$I=-\int(6x^2-6x)\log f(x)\mathrm dx.$$ Now from the functional equation we get $\log f(x)+\log f(1-x)=x^2-x.$ Substituting for $\log f(x)$ and splitting the integral now gives that $$I=-6\int_0^1(x^2-x)^2\mathrm dx+6\int(x^2-x)\log f(1-x)\mathrm dx=-1/5+6\int(x^2-x)\log f(x)\mathrm dx,$$ by using the property $\int_a^bg(x)\mathrm dx=\int_a^bg(a+b-x)\mathrm dx$ and simplifying. Integrating the second part of $I$ by parts now gives that $$I=-1/5+6\log f(x)\int \mathrm d\left(\frac{2x^3-3x^2}{6}\right)-\int_0^1\frac{(2x^3-3x^2)f'(x)}{f(x)}\mathrm d x=-1/5-I,$$ giving us that $$I=-1/10.$$
