$$ \int_0^{\infty} xe^{-x(y+1)}dx$$ $$=-\frac x {y+1}e^{-x(y+1)}|_0^{\infty} +\frac 1 {y+1} \int_0^{\infty}e^{-x(y+1)}dx$$ $$=\frac 1{(y+1)^{2}}.$$
Unfortunately i cannot follow the steps. I guess integration by parts has been used?
How can Gamma density approach be applied here?
by parts, with $a=y+1$
$u=x$ and $v'=e^{-ax}$.
$$\int_0^Xxe^{-ax}dx=$$
$$\Bigl[x\frac{-1}{a}e^{-ax}\Bigr]_0^X-\int_0^X\frac{-1}{a}e^{-ax}dx$$
$$\frac{Xe^{-aX}}{a}+\frac{1}{a}\Bigl[\frac{-1}{a}e^{-ax}\Bigr]_0^X=$$
$$\frac{Xe^{-aX}}{a}+\frac{1}{a^2}-\frac{e^{-aX}}{a^2}$$
now take the limit when $X\to+\infty$.