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

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 $$4 \left[\left(\tan^{-1} \sqrt{\frac{y}{4-y}} \right)^2\right]_{y=0}^1 = \frac{\pi^2}{9}.$$

$$I=\int_0^1 \frac{\ln(1-x+x^2)}{x(x-1)}dx=-\int_0^1 \frac{\ln(1-x(1-x))}{x}dx-\int_0^1\frac{\ln(1-x(x-1))}{1-x}dx$$ 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).