Problem integrating when attempting a solution with the Poincaré Lemma 
d) This is part I am having troubles with. 
I get that
$$
\hat{\mathbb{X}}_t = \left(\frac{\partial}{\partial t}\hat{\Phi}_t \right) \Phi_t^{-1} 
= \left(\frac{\partial}{\partial t}\hat{\Phi}_t \right) (x/t,y/t,z) 
= (x/t,y/t,0)
$$
so,
$$\begin{align}
i_{\hat{\mathbb{X}}_t}\beta &= i_{\hat{\mathbb{X}}_t} \left[\frac{3xz}{r^5} dy\wedge dz + \frac{3yz}{r^5} dz\wedge dx +\frac{2z^2-x^2-y^2}{r^5}dx \wedge dy \right] \\
&= \frac{3xz}{r^5}\left[\frac{y}{t}dz\right]+\frac{3yz}{r^5}\left[\frac{-x}{t}dz\right]+\frac{2z^2-x^2-y^2}{r^5}\left[\frac{x}{t}dy-\frac{y}{t}dx\right] \\
&= \frac{2z^2-x^2-y^2}{r^5}\left[\frac{x}{t}dy-\frac{y}{t}dx\right]
\end{align}$$
Then 
and so
$$\begin{align}
\Phi_t^*\left(i_{\hat{\mathbb{X}}_t}\beta\right) &= \frac{2z^2-t^2x^2-t^2y^2}{(t^2x^2+t^2y^2+z^2)^{5/2}}\left[txdy-tydx \right] \\
&= \frac{2z^2-t^2x^2-t^2y^2}{(t^2x^2+t^2y^2+z^2)^{5/2}}txdy - \frac{2z^2-t^2x^2-t^2y^2}{(t^2x^2+t^2y^2+z^2)^{5/2}}tydx
\end{align}$$
However I cannot see how to integrate this. Help much appreciated.
UPDATE: Apparently the correct answer is $\alpha=(x dy -y dx)/r^3$. This would suggest I am on the right track.
The following integral may help $\displaystyle \int_0^w \frac{v^n}{(1+v)^s}dv=\frac{1}{1-s}\frac{w^n}{(1+w)^{s-1}}-\frac{n}{1-s}\int_0^w \frac{v^{n-1}}{(1+v)^{s-1}}dv$ $n>0, s\neq 1,w > -1$
I believe I may be missing something as I have had a number of problems with these type of questions. See my other question on the Poincaré Lemma application.

Let $$I(x,y,z) = \int_0^1 \frac{2z^2 - t^2x^2-t^2y^2}{(t^2x^2+t^2y^2+z^2)^{5/2}}t\, dt.$$
Using the $u$-substitution $u = t^2x^2 + t^2y^2 + z^2$, we compute
\begin{align}I(x,y,z) &= \frac{1}{2(x^2+y^2)}\int_{z^2}^{r^2} \frac{2z^2 - (u - z^2)}{u^{5/2}}\, du\\
&= \frac{1}{2(x^2 + y^2)} \int_{z^2}^{r^2} \left(\frac{3z^2}{u^{5/2}} - \frac{1}{u^{3/2}}\right)\, du\\
&= \frac{1}{2(x^2 + y^2)} \left\{-\frac{2z^2}{u^{3/2}}\bigg|_{u = z^2}^{u = r^2} + \frac{2}{u^{1/2}}\bigg|_{u = z^2}^{u = r^2}\right\}\\
&= \frac{1}{2(x^2 + y^2)} \left\{\frac{2}{z} - \frac{2z^2}{r^3} + \frac{2}{r} - \frac{2}{z}\right\}\\
&= \frac1{2(x^2 + y^2)}\frac{2r^2 - 2z^2}{r^3}\\
&= \frac1{2(x^2 + y^2)}\frac{2(x^2 + y^2)}{r^3}\\
&= \frac{1}{r^3}.
\end{align}
Therefore, $$\alpha = I(x,y,z)(x\, dy - y\, dx) = \frac{x\, dy - y\, dx}{r^3}.$$
