Find the integral from 31/2 to 3 of 1/(x) ((x2 +9))1/2
I tried to substitute x= 3tanu, and I get down eventually to a solution that is (sec(pi/4) - sec(pi/6))
I realize integrals can have many forms, but the form in the back of my book isnt even close to sec(arctan(3/3)-sec(arctan(31/2/3). My answer doesnt match the books. Maybe its a domain problem?
The books answer is: 1/6ln((3-2*21/2)(7+4*31/2
Sorry, I'm exhausted, and I do not have the energy to type out how I got from the tangent substitution to my final value.
Any help would be great.