Answer
If $f_Y(y) = (n+2)(n+1)(1-y)^ny,\ 0\le y\le 1$, then a) $\color{blue}{f_y(y)\ge 0}$ and b) $\color{blue}{\int_{-\infty}^\infty f_Y(y)\ dy =1}$ (see the details below)
Work Step by Step
To show that $f_Y(y) = (n+2)(n+1)y^n(1-y),\ 0\le y\le 1$ is a pdf it suffices to show that: a) $f_Y(y) \ge 0$ and b) $\int_{-\infty}^\infty f_Y(y)\ dy = 1$ (see Example 3.4.5, p. 134).
$\underline{\text{a) Show}\ f_Y(y) \ge 0}$
$\begin{align*}
0\le y\le 1 &\implies (y\ge 0)\wedge(1\ge y) \\
&\implies (y\ge 0) \wedge ( 1-y \ge 0) \\
&\implies (y\ge 0) \wedge ( (1-y)^n \ge 0,\ n\in \mathbb{Z}^+) \\
&\implies (n+2)(n+1)(1-y)^ny \ge 0,\ n\in \mathbb{Z}^+ \\
&\implies (n+2)(n+1)(1-y)^ny \ge 0,\ 0\le y\le 1,\ n\in\mathbb{Z}^+ \\
0\le y\le 1 &\implies \color{blue}{f_Y(y) \ge 0}
\end{align*}$
$\underline{\text{b) Show}\ \int_{-\infty}^{\infty} f_Y(y) =1}$
$\begin{align*}
\int_{-\infty}^\infty f_Y(y)\ dy &= \int_{-\infty}^0 f_Y(y)\ dy + \int_0^1 f_Y(y)\ dy + \int_1^\infty f_Y(y)\ dy \\
&= \int_{-\infty}^0 0\ dy + \underbrace{\int_0^1 (n+1)(n+1)(1-y)^ny\ dy}_{\begin{array}{l}\text{Let}\ u\ =\ 1-y \implies (du = -dy)\wedge (y=1-u) \\ \text{Also},\ (y=0 \implies u=1)\wedge(y=1 \implies u=0)\end{array}} + \int_1^\infty 0\ dy \\
&= 0 + (n+2)(n+1)\int_1^0 u^n(1-u)\ (-du) + 0 \\
&= (n+2)(n+1)\int_0^1 u^n(1-u)\ du \\
&= (n+2)(n+1)\int_0^1 (u^n -u^{n+1})\ du \\
&= (n+2)(n+1)\left( \frac{u^{n+1}}{n+1} - \frac{u^{n+2}}{n+2} \right]_0^1 \\
&= (n+2)(n+1)\left[ \left(\frac{1^{n+1}}{n+1} - \frac{1^{n+2}}{n+2}\right) - \left(\frac{0^{n+1}}{n+1} - \frac{0^{n+2}}{n+2}\right) \right] \\
&= (n+2)(n+1)\left[ \left(\frac{1}{n+1} - \frac{1}{n+2}\right) - \left(0 - 0\right) \right] \\
&= \frac{(n+2)(n+1)}{n+1} - \frac{(n+2)(n+1)}{n+2} \\
&= (n+2) - (n+1) \\
&= n-n + 2-1 \\
\color{blue}{\int_{-\infty}^\infty f_Y(y)\ dy} &\color{blue}{= 1}
\end{align*}$
