1.) The problem is: Find the arc length of f(x)= x^3/3-1/(4x) from x=1 to 2 2.) Relevant formulas: ds = √(1+(dy/dx)) abs(L) = ∫ds 3.) My work so far: f'(x)= x^2+1/(4x^2) abs(L) = ∫(from 1 to 2) √(1+(x^2+1/(4x^2))^2 dx = ∫(from 1 to 2) √(1+(x^4+1/2+1/(16x^4)) dx = ∫(from 1 to 2) √(x^4+3/2+1/(16x^4)) dx = ∫(from 1 to 2) √((16x^8+24x^4+1)/(16x^4) dx = 1/4∫(from 1 to 2) √((16x^8+24x^4+1)/(x^4) dx = 1/4∫(from 1 to 2) √(16x^8+24x^4+1)/(x^2) dx Generally, I would suspect the numerator would be able to be factored and that would make this integral much easier to solve. I'm stumped after this part, because it seems that the numerator is irreducible. I'm up for any help, I'm not asking for the answer. I really just need guidance from this point...that includes the possibility that the professor may have made a typo. I really wanted to consider that I may just not be aware of the next step before accusing the professor of fat fingering this. Thanks for the help.