1. The problem statement, all variables and given/known data Integral of dt/ (√(t^2 -6t + 13) 2. Relevant equations I sub v = t-3, and v = 2tan(θ) 3. The attempt at a solution first I completed the square of t^2 -6t + 13, I got (t-3)^2 +4 Also I say v = t-3 ∫ dv/ (√(v^2 +4) I then sub in trig ∫(2sec^2(θ)dθ / 2√(tan^2(θ) +1) = ∫ (sec^2(θ)/ secθ)dθ = ln|secθ +tanθ| = ln|(√v^2+4)/2 + v/2| edit sec here = ln| √((t-3)^2 +4/2) + (t-3)/2| +c I feel it isn't correct. What did I do?