Integral of 6x3 32x2 78x 18 x2 x2 6x 18Solution the pro

Solution

the procedure is correct. change the constants to get correct result . ThanQ x^2-6x+8=x^2-6x+9-1 x^2-6x+8=(x-3)^2-1 Let: x-3=sec(u) Leaving: x^2-6x+8=sec^2(u)-1 And because: sec^2(u)-1=tan^2(u) You have: S cot^2(u) dx Now if: x-3=sec(u) Then; dx=sec(u)tan(u)du Leaving: S cot^2(u) sec(u)tan(u)du And because: cot^2(u)*tan(u)=cot(u) You have: S cot(u)sec(u) du And because: cot(u)*sec(u)=csc(u) You have: S csc(u) du Integrating: -ln(csc(u)+cot(u)) And plugging in: u=arcsec(x-3) You are done. (Note that although this looks crazy,if you graph it, this answer is identical to the one you had)