Peskin & Schroeder's way of showing $Z_1=Z_2$ via integration by parts 
I am trying to follow Peskin & Schroeder's textbook on Renormalization.  I tried a few ways but this does not match with the textbook.
First equation (10.43 )in Peskin is given
\begin{align}
  \delta_2 = -\frac{e^2}{(4\pi)^{\frac{d}{2}}}
  \int_0^1 dx \frac{\Gamma\left(2-\frac{d}{2}\right)}{\left( (1-x)^2 m^2 + x
  \mu^2 \right)^{2-\frac{d}{2}}}
  \left[ (2-\epsilon) x - \frac{\epsilon}{2} \frac{2x(1-x)m^2}{\left( (1-x)^2
  m^2 + x \mu^2 \right)} (4-2x - \epsilon (1-x)) \right]. \label{1043}
\end{align}
and equation (10.46) in Peskin is given
\begin{align}
  \delta_1 &= -\frac{e^2}{(4\pi)^{\frac{d}{2}}} 
  \int_0^1 dz (1-z)   \\
  &\left\{ 
  \frac{\Gamma\left(2-\frac{d}{2}\right)}{\left( (1-z)^2 m^2 + z \mu^2 \right)^{2-\frac{d}{2}}} \frac{(2-\epsilon)^2}{2} 
  + \frac{\Gamma \left(3-\frac{d}{2}\right)}{\left( (1-z)^2 m^2 + z \mu^2 \right)^{3-\frac{d}{2}}} \left( 2 (1-4z + z^2) - \epsilon(1-z)^2 \right) m^2 \right\}. \label{1046}
\end{align}
From integration by parts I want to obtain 10.46 to 10.43

My first trial was re-write equation 10.46 as
\begin{align}
\delta_1 = -\frac{e^2}{(4\pi)^{\frac{d}{2}}} \int_0^1 dz(1-z) 
\frac{\Gamma(2-\frac{d}{2})}{((1-z)^2 m^2 + z \mu^2)^{2-\frac{d}{2}}} 
\left[ \frac{(2-\epsilon)^2}{2} + \frac{(2-\frac{d}{2})}{((1-z)^2 m^2 + z \mu^2)} (2(1-4z+z^2) -\epsilon (1-z)^2 ) m^2\right] 
\end{align}
and then do integration by parts.  [replacing $(1-z) \rightarrow x$ is not a good choice]
First I just compute with mathematica and later i noticed that I have a problem with boundary term.
Do you have any ideas?
 A: First equation (10.43 )in Peskin is given
\begin{align}
  \delta_2 = -\frac{e^2}{(4\pi)^{\frac{d}{2}}}
  \int_0^1 dx \frac{\Gamma\left(2-\frac{d}{2}\right)}{\left( (1-x)^2 m^2 + x
  \mu^2 \right)^{2-\frac{d}{2}}}
  \left[ (2-\epsilon) x - \frac{\epsilon}{2} \frac{2x(1-x)m^2}{\left( (1-x)^2
  m^2 + x \mu^2 \right)} (4-2x - \epsilon (1-x)) \right]. 
\end{align}
and equation (10.46) in Peskin is given
\begin{align}
  \delta_1 &= -\frac{e^2}{(4\pi)^{\frac{d}{2}}} 
  \int_0^1 dz (1-z)   \\
  &\left\{ 
  \frac{\Gamma\left(2-\frac{d}{2}\right)}{\left( (1-z)^2 m^2 + z \mu^2 \right)^{2-\frac{d}{2}}} \frac{(2-\epsilon)^2}{2} 
  + \frac{\Gamma \left(3-\frac{d}{2}\right)}{\left( (1-z)^2 m^2 + z \mu^2 \right)^{3-\frac{d}{2}}} \left( 2 (1-4z + z^2) - \epsilon(1-z)^2 \right) m^2 \right\}.  
\end{align}
Want to show from 10.46 to 10.43 using integration by parts.
Using
\begin{align}
  &\frac{d}{dz}\left[ \frac{\Gamma\left(2-\frac{d}{2}\right)}{\left( (1-z)^2 m^2 + z \mu^2 \right)^{ 2-\frac{d}{2}}} \right] =  \frac{\Gamma\left(3-\frac{d}{2}\right)}{\left( (1-z)^2 m^2 + z \mu^2 \right)^{ 3-\frac{d}{2}}} \left( 2m^2(1-z) - \mu^2 \right).  
\end{align}
Now we subtract $\delta_1$ and $\delta_2$ and collect (1-2z). For the (1-2z) terms replace this by total-derivatives, we have
\begin{align}
   \delta_1 - \delta_2 
   &\equiv -\frac{\epsilon}{2}\frac{e^2}{(4\pi)^{\frac{d}{2}}} 
   \int_0^1 dz (1-z) \frac{\Gamma\left(2-\frac{d}{2}\right)}{\left((1-z)^2 m^2 + z \mu^2\right)^{3-\frac{d}{2}}} \left( 2 m^2 (1-z)(1+ z(2-\epsilon) )- z \mu^2 (1-\epsilon) \right) .
 \end{align}
So at this moment, we see that finite parts of $\delta_1$ and $\delta_2$ coincides. i.e., In the limit $\epsilon \rightarrow 0$, $\delta_1 -\delta_2 \rightarrow 0$.
