Two independent random variable $X,Y$ are distributed on $[0,\infty)$ according to the cumulative distribution function $F(x)=1-(x+1)^{-2}$. Let $Z=\min(X,Y)$. Determine $E\left[\frac{Z}{Z+2}\right].$
By definition of expectation, we have $E\left[\frac{Z}{Z+2}\right]=\int_0^\infty\frac{z}{z+2}f(z)dz$, where $f(z)$ is the distribution function of $Z$. The distribution function of $X$ can be found by taking the derivative of $1-(x+1)^{-2}$, which gives $2(x+1)^{-3}$, but how can we find the distribution function of $Z$?
By definition:
$$\begin{align} F_Z(z) & = \mathsf P(\min \{X,Y\}\leq z) \\[1ex] &= 1-\mathsf P(\min\{X,Y\}>z) \\[1ex] &= 1-\mathsf P(X>z)\mathsf P(Y>z) \\[1ex] & = 1-(1-F(z))^2 \\[1ex] & = 1 - (z+1)^{-4} \end{align}$$
You can take it from here.