To approach this, I would apply the [binomial theorem][1], which holds for non-negative integer $c$:

$$
(a+b)^c = \sum_{i=0}^c {c\choose i} a^ib^{c-i}
$$

When you apply this identity to $(1-x)^{n-y}$, the integrand becomes a standard polynomial in $x$:

\begin{align*}
Pr(Y=y) &= {n\choose y}\int_0^1 \sum_{i=0}^{n-y} {n-y\choose i}(-1)^ix^{i+y}~dx \\
&= {n\choose y}\sum_{i=0}^{n-y}{n-y\choose i}(-1)^i\frac{x^{i+y+1}}{i+y+1}\bigg|_0^1 \\
&= {n\choose y}\sum_{i=0}^{n-y}\frac{{n-y\choose i}(-1)^i}{i+y+1}
\end{align*}


  [1]: https://en.wikipedia.org/wiki/Binomial_theorem