Computing the variance of a "complicated" random variable

Question

Let $f_{X}(x)=\frac{1}{(x+1)^2}$ when $x\geq 0$, and $0$ otherwise. I'm being asked to compute the following expectation: $\mathbb{E}[(1+X)^2e^{-2X}].$ Now, I have decomposed the expression in this way: $$\mathbb{E}[(1+X)^2e^{-2X}]= \mathbb{E}[X^2e^{-2X}]+\mathbb{E}[2Xe^{-2X}]+\mathbb{E}[e^{-2X}].$$ Assuming the multiplication of two random variables allow you to just multiply them in the definition of the expectation, I get complicated expressions that suggest to me I'm in the wrong track or there's something I'm not seeing. For example: $$\mathbb{E}[X^2e^{-2X}]=\int_{0}^{\infty}x^2 \left(\frac{1}{(x+1)^2}\right)e^{-\frac{2}{(1+x)^2}}.$$ The similarity with the normal distribution tells me that maybe it's part of the solution. But again, I'm not sure I'm even on the right track with this. Thank you.

Answer 1

Expanding $(1+X)^{2}$ is a bad idea.

$E[(1+X)^{2}e^{-2X}]=\int_0^{\infty} (1+x)^{2}\frac 1 {(1+x)^{2}}e^{-2x}dx=\int_0^{\infty} e^{-2x}dx=\frac 1 2$.