Solving Line Integral on Curve C

[etotheix]

Homework Statement

Evaluate the following line integral on the indicated curve C

$$\int(y^2-x^2)ds$$

C: x = 3t(1+t), y=t^3 ; 0 <= t <= 2

Homework Equations

ds = $$\sqrt{(f'(t))^2+(g'(t))^2}dt$$

The Attempt at a Solution

dx/dt = 3+6t
dy/dt = 3t^2

ds = $$\sqrt{(3+6t)^2+(3t^2)^2}$$dt
ds = 3*$$\sqrt{t^4+4t^2+4t+1}$$dt

y^2 = t^6
x^2 = (3t+3t^2)^2 = 9t^2+18t^3+9t^4

$$\int(y^2-x^2)ds = 3\int(t^6-9t^2-18t^3-9t^4)\sqrt{t^4+4t^2+4t+1}dt$$

From here I don't know how to solve the integral, and it looks way more complicated than what we have done so far in class. Maybe I am doing something wrong? Or there is a trick involved here? Any help would be greatly appreciated. Thanks.

 

[Mark44]

I see a typo, but I can't offer any more help than that.
$$\int(y^2-x^2)ds = 3\int(t^6 -9t^2-18t^3-9t^4)\sqrt{t^4+4t^2+4t+1}dt$$

 

[etotheix]

Thanks Mark44, I corrected the mistake in the first post. I will come back to this thread if I find an answer and post it.