Line integral over smooth curves

Let $C$ be the line segment from $(0,0)$ to $(1,1)$. Compute $\int_{C}{(3x^2y^2+1) \, dx + (2x^3y) \, dy}$.

I did this by setting $x=t$ which implied $y=t$ along $C$. Thus, the integral became

$$\int_0^1 (5t^5+1) \, dt=2$$

Now, if we choose some other smooth curve $C^'$ from $(0,0)$ to $(1,1)$, then what can we say about the value of

$$\int_{C^'} (3x^2y^2+1) \, dx + (2x^3y) \, dy?$$

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It will be the same, because $\int_{C'}(3x^2y^2+1)dx+(2x^3y)dy = \int_{C'}d(x^3y^2+x)=x^3y^2+x\Bigl|_{(0,0)}^{(1,1)}=2$ – Norbert Aug 12 '12 at 15:19

Depending on the source you learned calculus from, there are lots of different ways to say and write this. Think of $\mathbf{v}(x,y)=(3x^2y^2 +1, \ 2x^3y)$ as a vector field. If $dr=(dx, dy)$, then the integral you wrote is $\int_C \mathbf{v}\cdot dr$.
It turns out that the value of the integral is independent of the path you take if the vector field is what is called conservative. This just means that you can find some function $V:\mathbb{R}^2\to \mathbb{R}$ such that the gradient $\nabla V=\mathbf{v}$ (called the potential of the vector field).
In your case take $V(x,y)=x^3y^2 + x$. You get $\displaystyle \nabla V=\left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}\right)=(3x^2y^2+1, \ 2x^3y)=\mathbf{v}$.
Thus the vector field is conservative and hence any path you pick starting at $(0,0)$ and ending at $(1,1)$ will give you the same answer.