Double Integration Problem $\int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$ 
Compute
  $$I = \int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$$

Here are my steps:
$$\begin{split}
I &=\int_{0}^{1} \left(\int_0^1 \frac{dy}{1+y(x^2-x)}\right)dx\\
  &=\int_{0}^{1} \left[\frac{\ln(1+y(x^2-x))}{x^2-x}\right]_0^1dx\\
  &=\int_{0}^{1} \left[\frac{\ln(1+(1)(x^2-x))}{x^2-x}
                      -\frac{\ln(1+(0)(x^2-x))}{x^2-x}\right]dx\\
  &=\int_{0}^{1} \frac{\ln(1+x^2-x)}{x^2-x}dx
\end{split}
$$
And here I can't find any substitution to solve this integral.
Can anyone help me?
By the way, I also used Simpson's 3/8 method to find the approximation and got $1.063$.
But I want to find it using Calculus.

Switching the order of integration, we get $$4 \int_{y=0}^1 \frac{1}{\sqrt{y(4-y)}} \tan^{-1} \sqrt{\frac{y}{4-y}}\, dy.$$  Then the substitution $$u = \tan^{-1} \sqrt{y/(4-y)}, \quad du = \frac{1}{1+\frac{y}{4-y}} \cdot \frac{2}{(4-y) \sqrt{y(4-y)}} \, dy = \frac{1}{2 \sqrt{y (4-y)}} \, dy,$$ gives $$4 \left[\left(\tan^{-1} \sqrt{\frac{y}{4-y}} \right)^2\right]_{y=0}^1 = \frac{\pi^2}{9}.$$
