Solve For x?

Firstly, we need $\ log_{6} ( \frac{x^2 -x}{x^2 +1}\ )$ to be defined so we need the input to be more than zero.... $\ x(x-1) > 0 \Rightarrow\ x < 0 \ or \ x > 1$ . Then we need the next log to be defined such that $\ log_{6} ( \frac{x^2 -x}{x^2 +1}\ ) > 0$ .... $\frac{x^2 -x}{x^2 +1} > 1 \Rightarrow\ x^2 -x > x^2 +1 \therefore\ x < -1$ . For the last part we need $\ log_{0.2} (y) < log_{0.2} (1)$ which implies that $\ log_{6} ( \frac{x^2 -x}{x^2 +1}\ ) < 1 = log_{6} (6)$ ... $\frac{x^2 -x}{x^2 +1} < 6 \Rightarrow\ 5x^2 +x+6 > 0$ This is true for all real x (sketch a graph and find the minimum point or show that the expression never equals zero ). $\therefore\ answer = (- \infty , -1)$