By completing the square,
t^2-6t+13=(t^2-6t+9)+4=(t-3)^2+2^2
So, we can rewrite the integral as
int{dt}/{sqrt{(t-3)^2+2^2}}
Let t-3=2tan theta.
Rightarrow {dt}/{d theta}=2sec^2 theta
Rightarrow dt=2sec^2 theta d theta
by the above substitution,
=int{2sec^2theta d theta}/{sqrt{(2tan theta)^2+2^2)}
=int{sec^2theta}/{sqrt{tan^2theta+1}}d theta
by the identity tan^2theta+1=sec^2theta,
=int{sec^2theta}/{sqrt{sec^2theta}}d theta=int sec theta d theta=ln|sec theta + tan theta|+C_1
by using
tan theta={t-3}/2 and sec theta=sqrt{tan^2theta+1}={sqrt{t^2-6+13}}/2,
we have
=ln|{sqrt{t^2-6+13}}/2+{t-3}/2|+C_1
=ln|{sqrt{t^2-6t+13}+t-3}/2|+C_1
by log property,
=ln|sqrt{t^2-6t+13}+t-3|-ln2+C_1
by setting C=ln2+C_1,
=ln|sqrt{t^2-6t+13}+t-3|+C