I need a bit of help integrating
X^2/(X^2-4X+5)
$\displaystyle \int \frac{x^2}{x^2-4x+5} dx = \int 1+ \frac{4x-5}{x^2-4x+5} dx = \int 1 dx + \int \frac{4x-5}{x^2-4x+5} dx$
$\displaystyle = \int 1 dx + \int \frac{4x-8}{x^2-4x+5} dx + \int \frac{3}{x^2-4x+5} dx$
$\displaystyle =x + 2ln |x^2-4x+5| + \int \frac{3}{x^2-4x+5} dx$
Can you continue now?
Look at the reply by Also sprach Zarathustra
$\displaystyle \int \frac{x^2}{x^2-4x+5}dx= \int \frac{4x-8}{x^2-4x+5}dx+\int \frac{3}{x^2-4x+5}dx + \int 1 dx$
use the substitution rule for the first term only:
$\displaystyle \int \frac{4x-8}{x^2-4x+5}dx = \int \frac{2(x-4)}{x^2-4x+5}$
sub. rule gives you:
$\displaystyle \int \frac{2}{u} du$ . complete it..
For the second term: $\displaystyle \int \frac{3}{x^2-4x+5}dx = \int \frac{3}{1+(x-2)^{2}} dx$
you should be able to finish it off now..