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#
