I complete the square for let (x-3) = 2 sec theta and dx = 2 sec theta tan theta dtheta
is this correct? wolfram did it differently so I just want to check to see if my way is correct as well.
your way looks correct as well. The plots of your function and wolfram's integral lie on top of each other.
I have to say though that this is clearly a partial fractions problem and not a trig substitution problem even if you can find an substitution that works.
$x^2-6x+5=(x-5)(x-1)$ so a bit of work shows
$\dfrac{1}{x^2-6x+5}=\dfrac{1}{4}\left(\dfrac{1}{x-5}-\dfrac{1}{x-1}\right)$
and the integral pops right out.