Integrating w.r.t. two different variables I can't really follow what's going on here to get from line 2 to line 3:
\begin{align}
& \int^\infty_0 \frac{d^2 p}{(2\pi)^2} \frac{\Lambda^2}{(p^2+\Lambda^2)(p^2+m^2)} \\[10pt]
= {} & \int^1_0 dz \, \frac{d^2 p}{(2\pi)^2}\frac{\Lambda^2}{(p^2+z\Lambda^2+(1-z) m^2)^2} \\[10pt]
= {} & \int^1_0dz \, \frac{1}{4\pi}\frac{\Lambda^2}{z\Lambda^2+(1-z)m^2} \\[10pt]
= {} & \frac{1}{4\pi}\frac{\Lambda^2}{\Lambda^2-m^2}\ln\frac{\Lambda^2}{m^2}
\end{align}
There are two different integration variables and then one of them vanishes. I've tried evaluating the integral w.r.t $p$ and then again w.r.t. $p$ because it's $d^2 p$, but that didn't work (just using wolfram alpha).  From line one to line two I can follow, it's a Feynman parametrization, but what do you do to get from line 2 to line 3? 

The integral over momentum space is missing in Line 2 of the OP.  That is to say that the expression $\displaystyle d^2 p\,\frac{\Lambda^2}{(p^2+z\Lambda^2+(1-z) m^2)^2}$ should read
$$\int_{-\infty}^\infty\int_{-\infty}^\infty \frac{\Lambda^2}{(p_1^2+p_2^2+z\Lambda^2+(1-z) m^2)^2}\,dp_1\,dp_2$$
which after a transformation to polar coordinates $(p_1,p_2)\mapsto(\rho,\phi)$ becomes
$$\begin{align}
\int_0^\infty\int_0^\infty \frac{\Lambda^2}{(p_1^2+p_2^2+z\Lambda^2+(1-z) m^2)^2}\,dp_1\,dp_2&= \int_0^{2\pi}\int_0^\infty \frac{\Lambda^2\rho}{(\rho^2+z\Lambda^2+(1-z) m^2)^2}\,\rho\,d\rho\,d\phi\\\\
&=2\pi \int_0^\infty \frac{\Lambda^2\rho}{(\rho^2+z\Lambda^2+(1-z) m^2)^2}\,\rho\,d\rho\\\\\
&=2\pi \left.\left(-\frac12 \frac{1}{\rho^2+z\Lambda^2+(1-z)m^2}\right)\right|_{\rho=0}^{\rho\to \infty}\\\\
&=2\pi\left(\Lambda^2 \frac12 \frac{1}{z\Lambda^2+(1-z)m^2}\right)
\end{align}$$
Finally, dividing by $4\pi^2$ and integrating over $z$ from $0$ to $1$ yields the coveted result.
