<s>[INST] 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.
 [/INST] $$I=\int_0^1 \frac{\ln(1-x+x^2)}{x(x-1)}dx=-\int_0^1 \frac{\ln(1-x(1-x))}{x}dx-\int_0^1\frac{\ln(1-x(x-1))}{1-x}dx$$
The substitution $1-x\to x $ in the second integral reveals that:
$$\int_0^1\frac{\ln(1-x(1-x))}{1-x}dx=\int_0^1\frac{\ln(1-x(1-x))}{x}dx$$
$$\Rightarrow I=-2\int_0^1\frac{\ln(1-x(1-x))}{x}dx=2\int_0^1\frac{\ln(1+x)-\ln(1+x^3)}{x}dx$$
But we are lucky again, because: $$\int_0^1\frac{\ln(1+x^3)}{x}dx\overset{x=t^{1/3}}=\frac13\int_0^1 \frac{\ln(1+t)}{t^{1/3}}\,t^{1/3-1}dt\overset{t=x}=\frac13\int_0^1\frac{\ln(1+x)}{x}dx$$
Also we can use this equality which will lead to:
$$ I=\frac43 \int_0^1 \frac{\ln(1+x)}{x}dx=-\frac23 \int_0^1 \frac{\ln x}{1-x}dx=-\frac23 \sum_{n=0}^\infty \int_0^1 x^n\ln x dx=\frac23 \sum_{n=1}^\infty \frac{1}{n^2}$$
Combining with heropup's answer we can deduce that $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$
I guess we can add one more to this big list Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem).
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