Say for instance you were given that $\mathbf r = t^3 \vec i+t^2\vec j+t\vec k$ and $\mathbf A = sinx \vec i+cosy \vec j+xz \vec k$.
When calculating the line integral $$\oint\mathbf A \bullet d\mathbf r$$ do you do anything different to when calculating $$\int\mathbf A \bullet d\mathbf r$$?
On
In general there is no difference. However, if you know that $\mathbf{A}$ is a conservative field, then $$\oint \mathbf{A} \cdot d\mathbf{r} = 0, \int_\gamma \mathbf{A} \cdot d\mathbf{r} = \mathbf{A}(\gamma(b)) - \mathbf{A}(\gamma(a))$$ where $\gamma(a),\gamma(b)$ are the starting and end points of $\gamma$, respectively.
As pointed out in the comments, the circle on the integral sign indicates the integral is taken over a closed curve in space; it can be omitted without ambiguity when $\mathbf{r}$ is given, and in some sense they are exactly the same notation, just one is slightly more specific to the example you are working with.
THAT SAID, if you have a closed curve (which in your example you do not), you may apply Stokes' Theorem, which sometimes makes the calculation simpler if you can find a simple enough surface whose boundary is the curve in question and the curl $\nabla\times\mathbf{A}$ is simpler than $\mathbf{A}$. If the curve is not closed, then Stokes' theorem makes no sense and the integral has to be computed the old-fashioned way...