Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 24

intdt/sqrt(t^2-6t+13)
Let's evaluate the integral by rewriting it by completing the square on the denominator,
=intdt/sqrt((t-3)^2+4)
Now let's use the integral substitution,
Let u=t-3
=>du=dt
=int(du)/sqrt(u^2+4)
Now use the trigonometric substitution: For sqrt(bx^2+a) substitute x=sqrt(a)/sqrt(b)tan(v)
So ,Let u=2tan(v)
=>du=2sec^2(v)dv
=int(2sec^2(v)dv)/sqrt((2tan(v))^2+4)
=int(2sec^2(v)dv)/sqrt(4tan^2(v)+4)
=int(2sec^2(v)dv)/sqrt(4(tan^2(v)+1))
=int(2sec^2(v)dv)/(2sqrt(tan^2(v)+1))
=int(sec^2(v)dv)/sqrt(tan^2(v)+1)
Now use the identity:1+tan^2(x)=sec^2(x)
=int(sec^2(v)dv)/sqrt(sec^2(v))
=intsec(v)dv
Now use the standard integral.
intsec(x)dx=ln|sec(x)+tan(x)|
=ln|sec(v)+tan(v)|
Substitute back tan(v)=u/2
=>1+tan^2(v)=sec^2(v)
=>1+(u/2)^2=sec^2(v)
=>sec^2(v)=(u^2+4)/4
=>sec(v)=sqrt(u^2+4)/2
Plug these into the solution, thus
=ln|sqrt(u^2+4)/2+u/2|
Now plug back u=t-3 and add a constant C to the solution,
=ln|sqrt((t-3)^2+4)/2+(t-3)/2|+C