We are told $k$ is an algebraically closed field of characteristic $2$. We write $f \in k[x]$ as $\displaystyle f(x) = \displaystyle\sum_{i = 0}^s a_i x^i$ and, since $k$ contains the square root of any of its elements, $f'(x)$ is a square, and can be written as $\left(\displaystyle\sum_{2j + 1 \leq s} \sqrt{a_{2j+1}}\ x^i\right)^2$. Furthermore, since $k$ algebraically closed we can write $f'(x) := \displaystyle\left(\prod_{i=1}^t (x-b_i)^{r_i}\right)^2$.
The claim I am wondering about is how to check that the elements.
$\alpha_i := \frac{\sqrt{f(b_i)} - \sqrt{f(x)}}{(x - b_i)^{r_i}}$
are all integral over $k[x]$.
Since $k$ contains the square roots of any of its elements, I suppose $\sqrt{f(b_i)}$ is just something in $k$, and that all the coefficients of $\sqrt{f(x)}$ are also in $k$. Still, I don't know how to think about the denominator and its coefficients and what the coefficients of this entire rational function end up being.
Also, how does $\{ 1, \alpha_1, ..., \alpha_t \}$ generate the integral closure $B$ over $k[x]$?
EDIT: Some info I forgot to add:
$1.$ $f$ should not have double roots in $\overline{k}$
$2.$ I believe this can be reduced to figuring out why $((x - b_i)^{r_i})^2$ divides each of $f(b_i)$ and $f(x)$.
EDIT $2$:
$snake$ $\in$ $boot$
$f(x) = x^3 + 1$ and $f'(x) = 3x^2 \equiv x^2$, just for an example. You get zero a double root, so we should be able to write $f'(x) = (x - 0)^2$ and $\alpha$ would just be $\frac{\sqrt{1} - \sqrt{x^3 + 1}}{x}$ And $x^2$ does not divide the square of either of these summands in the numerator individually, but it does divide the sum of their squares, $x^3 + 2 \equiv x^3$. So $2.$ in my first edit is a bit too specialized... one has to look at the sum sometimes... How can we then prove the sum of the squares is actually divisible by that root?