Calculus Arc Length Question

Find the exact length of the curve.
x=1+3t^2 y=4+2t^3 0(less than or equal to)t(less than or equal to)pi

I use u=t^2 + t^4 du=2t + 4t^3
Changing the integral from 0 to 1 To 0 to 2.
Ending up with 4u^3/2 from 0 to 2 after integrating.
The answer I come up with is 4 times the square root of 8.
The answer should be 4 times( the square root of 8) - 2.
I dont know if my algebra is wrong, but I'm definitely using the right formula for arc length.

x = 1 + 3t² ------> dx/dt = 6t
y = 4 + 2t³ ------> dy/dt = 6t²

You have: 0 < t < pi, but then you seem to integrate from t = 0 to t = 1
I'll assume the latter

L = ∫₀¹ √((dx/dt)² + (dy/dt)²) dt
L = ∫₀¹ √((6t)² + (6t²)²) dt
L = ∫₀¹ √(36t² + 36t⁴) dt
L = ∫₀¹ 6t √(1 + t²) dt
L = 2 (1 + t²)^(3/2) |₀¹
L = 2 ((2)^(3/2) - 1^(3/2))
L = 2 (√8 - 1)
L = 2*2√2 - 2
L = 4√2 - 2

Sorry, I don't get 4√8 - 2 either, but noticed that I factored 36t² out of square root, instead of just 36 (which would have left 6√(t² + t⁴), which is much more complicated to integrate than 6t√(1 + t²))

-- Ματπmφm --

Nevermind I got it. Thank you.

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int(sqrt((dx/dt)^2 + (dy/dt)^2) * dt) =>
int(sqrt((6t)^2 + (6t^2)^2) * dt) =>
int(sqrt(36t^2 + 36t^4) * dt) =>
int(6t * sqrt(1 + t^2) * dt =>
6 * int(t * sqrt(1 + t^2) * dt)

u = 1 + t^2
du = 2t * dt

6 * (1/2) * int(u^(1/2) * du) =>
3 * (2/3) * u^(3/2) + C =>
2 * u^(3/2) + C =>
2 * (1 + t^2)^(3/2) + C

From 0 to pi

2 * (1 + pi^2)^(3/2) - 2 =>
2 * ((1 + pi^2)^(3/2) - 1)