Solving Integral of dt/ (√(t^2 -6t + 13) with Trig Substitution

[Jbreezy]

Homework Statement

Integral of dt/ (√(t^2 -6t + 13)

Homework Equations

I sub
v = t-3, and v = 2tan(θ)

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?

 

[vela]

One thing you did was write $\sec\theta$ in terms of $v$ incorrectly.

 

[Jbreezy]

OK, I fixed it. I flipped it back it was OK on my paper just not when I rewrote it on the forum.

 

[vela]

Try differentiating your result and seeing if you recover the integrand. That's the easiest way to check your answer.