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.
