Second cohomology group of twice-punctured plane I am looking for pure calculus solution for computing the second cohomology class $H^2(\mathbb{R}^2-\{p , q\}, \mathbb{R})$ i.e. plane with two points removed. 
I know that every closed $2$-form is exact for $\mathbb{R}^2-\{p , q\}$, but I cannot find the explicit $1$-form for every $2$-form that under the exterior derivative produces the $2$-form. I want to "integrate" the  $2$-form and get the $1$-form by "d" operator. Note you have to somehow integrate "around" the points.
This is a problem from page 19 of Bott and Tu. I am trying to read the book on my own which would not be brought up in May if this was a Spring course.

The plane with two points removed retracts to a wedge of circles. Start at $p$ and gradually unfold , i.e., make the hole about $p$ larger. Do the same with $q$ simultaneously. Maybe doing this in a square first will help. Then, by homotopy invariance, the homology of  $\mathbb R^2 -$ {$p,q$} is that of a wedge of two circles. For the top homology alone, you just need to decide if $\mathbb R^2 -$ $p,q$ is orientable or not.
If you prefer something more computational, you may want to use the Mayer-Vietoris sequence, and it would help if you accept that the plane minus a point, equivalently, the plane minus a disk retracts to $S^1$, and you know the (co)homology of $\mathbb R^2$. I can only think of one decomposition for  using Mayer-Vietoris.
