Complete the square, then do a trigonometric substitution.
#sqrt(13+12x-x^2)=sqrt(49-36+12x-x^2)=sqrt(49-(x-6)^2)#
So, you want to evaluate #intsqrt(49-(x-6)^2)dx#.
It's going to get very messy, so I want to let #x-6=7u#. It is then easy to see that #(x-6)^2=49u^2# and #dx=7du#, so the integral becomes #int7sqrt(49-49u^2) du=7int7sqrt(1-u^2)du=49intsqrt(1-u^2)du#,
Now do a trigonometric substitution. I'll use #u=sintheta# which, of course, makes #du=costheta d theta#.
This makes the integral: #49intsqrt(1-sin^2theta)costhetad theta#.
This, in turn, becomes #49intcos^2thetad theta#. Now, use the power reduction to re-write #cos^2theta# as #1/2(1+cos2theta)#.
The problem has become:
Evaluate: #49/2int(1+cos2theta)d theta=49/2(theta+1/2sin2theta)+C#
Backfilling: #theta=sin^-1u# and
since #u=sintheta# implies #costheta=sqrt(1-u^2)#
#1/2sin2theta=1/2(2sinthetacostheta)=usqrt(1-u^2)#
Thus, the integral evaluates(in terms of #u#) leaving +C for later, to: #49/2(sin^-1u+usqrt(1-u^2))=1/2(49sin^-1u+49usqrt(1-u^2))=1/2(49sin^-1((x-6)/7)+49((x-6)/7)sqrt(1-((x-6)/7)^2))=1/2(49sin^-1((x-6)/7)+49((x-6)/7)sqrt(1-((x-6)/7)^2))=1/2(49sin^-1((x-6)/7)+(x-6)sqrt(49-(x-6)^2))=1/2(49sin^-1((x-6)/7)+(x-6)sqrt(13+12x-x^2))#
Oh, yeah! Don't forget to put #+C#.
