Prove a 3 parameter integral identity I have stumbled upon the following identity:

$$\int_0^1 \frac{(1-t)^c}{(1-z t)^b} dt=\int_0^1 \frac{(1+t)^{b-c-2}}{(1+(1-z) t)^b} dt+\int_0^1 \frac{t^c(1+t)^{b-c-2}}{(1-z+t)^b} dt$$

It appears to work for all $b \in \mathbb{R}$, $c \geq 0$ and $|z|<1$.
The identity is related to the integral representation of ${_2F_1}$, the hypergeometric function, however I don't have a nice proof of it.

How can we prove this identity, and are the conditions on the parameters I listed correct?

 A: We want to show
\begin{eqnarray*}
\int_0^1 \frac{(1-t)^c}{(1-z t)^b} dt=\int_0^1 \frac{(1+t)^{b-c-2}}{(1+(1-z) t)^b} dt+\int_0^1 \frac{t^c(1+t)^{b-c-2}}{(1-z+t)^b} dt.
\end{eqnarray*}
Make the substitution $t=\frac{1}{u}$ into the third integral ($dt=-\frac{du}{u^2}$)
\begin{eqnarray*}
\int_0^1 \frac{t^c(1+t)^{b-c-2}}{(1-z+t)^b} dt = -\int_{\infty}^1 \frac{(1+\frac{1}{u})^{b-c-2}}{u^c(1-z+\frac{1}{u})^b} \frac{du}{u^2} \\
=\int_1^{\infty} \frac{(1+u)^{b-c-2}}{(1+(1-z) u)^b} du.
\end{eqnarray*}
Now combine this with the second integral & the RHS becomes 
\begin{eqnarray*}
\int_0^{\infty} \frac{(1+t)^{b-c-2}}{(1+(1-z) t)^b} dt.
\end{eqnarray*}
Now make the substition $w=\frac{t}{1+t}$ (so $dw=\frac{dt}{(1+t)^2}$ and $t=\frac{w}{1-w}$) and we get the LHS.
\begin{eqnarray*}
\int_0^{\infty} \frac{(1+t)^{b-c}}{(1+(1-z) t)^b} \frac{dt}{(1+t)^2} &=& \int_0^{1} \frac{(1+\frac{w}{1-w})^{b}}{(1+(1-z) \frac{w}{1-w})^b} \frac{1}{(1+\frac{w}{1-w})^c}dw \\
&=&\int_0^1 \frac{(1-w)^c}{(1-z w)^b} dw.
\end{eqnarray*}
