1. The problem statement, all variables and given/known data Find the arc length of the function f(x) = x(sqrt(x/2-x) from 0 to x Integral forms involving a+bu --> integral [sqrt(a+bu)]/u^2du 2. Relevant equations Arc Length = integral sqrt(1+(f'(x))^2)dx 3. The attempt at a solution First, I took the derrivate f'(x) = sqrt(x/2-x) + x/[(2-x)^2(sqrt(x/2-x))] = [x(2-x)+x]/[(2-x)^2(sqrt(x/2-x))] =[x(3-x)]/[(2-x)^2(sqrt(x/2-x)] Then, the square of f'(x) = [x^2(3-x)^2]/[x(2-x)^3] =[x(3-x)^2]/[(2-x)^3] Then I added 1 to the square of the derivative, which simplified to =[8-3x]/[2-x]^3 Then, the square root of the above =sqrt([8-3x]/[2-x]^3) I substituted 2-x = t =sqrt[(3t+2)/(t^3)] <--multiplied top and bottom by sqrt[t] =sqrt(3t^2+2t)/t^2 I'm stuck here..I'm allowed to use formulas from a table of integrals (i.e. the one specified above), but I can't seem to get it into that form ..