The random variable $\frac{1}{1 + X}$ takes on values $\frac{1}{1 + 1} = \frac{1}{2}$ and $\frac{1}{1 + 0} = 1$, each with probability $1 / 2$. Using result
R5274: Probability mass partition for expectation of a random natural number, we have
\begin{equation}
\begin{split}
\mathbb{E} \left( \frac{1}{1 + X} \right)
& = \frac{1}{1 + 1} \mathbb{P}(X = 1) + \frac{1}{1 + 0} \mathbb{P}(X = 1) \\
& = \frac{1}{2} \frac{1}{2} + \frac{1}{2} \\
& = \frac{1}{4} + \frac{2}{4} \\
& = \frac{3}{4}
\end{split}
\end{equation}
while, using result
R5283: Expectation of a standard bernoulli random boolean number, we know that
\begin{equation}
\mathbb{E}(1 + X)
= 1 + \mathbb{E} X
= 1 + \frac{1}{2}
= \frac{3}{2}
\end{equation}
and therefore
\begin{equation}
\frac{1}{\mathbb{E}(1 + X)}
= \frac{1}{3 / 2}
= \frac{2}{3}
\end{equation}
$\square$
