To approach this, I would apply the binomial theorem, 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 integral 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*}

To approach this, I would apply the binomial theorem, 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*}

I would approach this by expanding the polynomial $(1-x)^{n-y}$ and then integrating the resulting polynomial in $x$ as usual:

\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*}

To approach this, I would apply the binomial theorem, 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 integral 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*}