$\displaystyle \int \frac{1}{x^2-6x+5}dx $ I complete the square for $\displaystyle \int \frac{1}{(x-3)^2-4}dx$ let (x-3) = 2 sec theta and dx = 2 sec theta tan theta dtheta

$\displaystyle \int \frac{2 sec\theta tan \theta d\theta}{4(sec^2 \theta -1} $ $\displaystyle \Rightarrow \frac{1}{2} \int \frac {sec \theta}{tan \theta} d\theta$

$\displaystyle \Rightarrow \frac{1}{2} \int csc \theta d\theta = \frac{1}{2} ln |csc \theta - cot \theta| + C $

$\displaystyle = \frac{1}{2} ln| \frac{x-3}{\sqrt{(x-3)^2 -4}} - \frac{2}{\sqrt{(x-3)^2 -4}}|+C$

is this correct? wolfram did it differently so I just want to check to see if my way is correct as well.