Calculus: Find the area of the surface generated by....

x=t^2+1/(2t) y=4t^(1/2), for 1/sqrt2≤t≤1 about the y axis
i know the formula is
∫ (a to b) 2π x sqrt((dx/dt)^2+(dy/dt)^2)
i plugged in the stuff and i found the integral to be hard to solve can some one point me in the right direction for solving this integral

thank you so much

SA = 2π ∫ x ds
SA = 2π ∫ [t^2 + (1/2)t^(-1)] ds from 1/sqrt(2) to 1

ds = sqrt((dx/dt)^2 + (dy/dt)^2) dt

dx/dt = 2t - (1/2)t^(-2)
dy/dt = 2t^(-1/2)

(dx/dt)^2 = [2t - (1/2)t^(-2)]^2 = (1/4)t^(-4) + 4t^2 - 2t^(-1)
(dy/dt)^2 = [2t^(-1/2)]^2 = 4t^(-1)

(dx/dt)^2 + (dy/dt)^2 = (1/4)t^(-4) + 4t^2 - 2t^(-1) + 4t^(-1)
(dx/dt)^2 + (dy/dt)^2 = (1/4)t^(-4) + 4t^2 + 2t^(-1)
(dx/dt)^2 + (dy/dt)^2 = [(4t^3 + 1) / (2t^2)]^2 (*)
sqrt((dx/dt)^2 + (dy/dt)^2) = (4t^3 + 1) / (2t^2)
ds = (4t^3 + 1) / (2t^2) dt

Sub that in:

SA = 2π ∫ [t^2 + (1/2)t^(-1)] (4t^3 + 1) / (2t^2) dt from 1/sqrt(2) to 1
SA = 2π ∫ [2t^3 + (1/4)t^(-3) + 3/2] dt from 1/sqrt(2) to 1 (**)
SA = 2π[(1/2)t^4 - (1/8)t^(-2) + (3/2)t] 1/sqrt(2) to 1
SA = 2π[(1/2)(1)^4 - (1/8)(1)^(-2) + (3/2)(1)] - 2π[(1/2)(1/sqrt(2))^4 - (1/8)(1/sqrt(2))^(-2) + (3/2)(1/sqrt(2))]
SA = 2π[15/8] - 2π[(1/8)(6sqrt(2) - 1)]
SA = π[4 - (3/2)sqrt(2)]
SA ≈ 5.902

Done!

(*) You're probably wondering where that came from. If you factor out a 1/(4t^4), you'll get something that becomes a perfect square, and then 1/(4t^4) = 1/(2t^2)^2.

(**) This part is just a lot of tedious multiplication and long division.