How do you prove that [math]\frac{\mathrm{d} }{\mathrm{d} x} \frac{x}{1-x} = \frac{\mathrm{d} }{\mathrm{d} x} \frac{1}{1-x} = \frac{1}{(1-x)^2}[/math]?
How do I find [math] \int_0^1 \frac{1-x}{1+x+x^2} \, \mathrm{d}x [/math]?