Answer
The integral converges and its value is $\frac{1}{ 2}$.
Work Step by Step
We have
$$
\int_{0}^{\infty} \frac{d x}{(x+2)^{2}}= \int_{0}^{\infty} (x+2)^{-2} d(x+2)\\
=-(x+2)^{-1}|_0^{\infty}=-\frac{1}{x+2}|_0^{\infty}=-\frac{1}{\infty}+\frac{1}{0+2}=\frac{1}{ 2}
$$
Hence, the integral converges and its value is $\frac{1}{ 2}$.
