`int (dt)/sqrt(t^2 - 6t + 13)` Evaluate the integral

`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`