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

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 $$I=-1/10.$$
