These are 2 problems that I'm stumped on and can't get...we are supposed to evaluate the integral using trig substitution.
x^2/(3+4x-4x^2)
(x^2+1)/(x^2-2x+2)
$\displaystyle \frac{x^2 + 1}{x^2 - 2x + 2} = \frac{(x^2 - 2x + 2) + 2x - 1}{x^2 - 2x + 2} = 1 + \frac{2x - 1}{x^2 - 2x + 2}$
$\displaystyle = 1 + \frac{(2x - 2) + 1}{x^2 - 2x + 2} = 1 + \frac{2x - 2}{x^2 - 2x + 2} + \frac{1}{x^2 - 2x + 2}$
$\displaystyle 1 + \frac{2x - 2}{x^2 - 2x + 2} + \frac{1}{(x - 1)^2 + 1}$.
The first two terms are trivial to integrate. To integrate the last term make the substitution $\displaystyle x-1 = tan \theta$ (or just recognise the standard form).