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## 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?

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