What is the distribution of 1/($X$+1)? I have a problem that I'm having trouble figuring out the distribution with given condition.
It is given that 1/($X$+1), where $X$ is an exponentially distributed random variable with parameter 1.

Original Problem:
What is the distribution of 1/($X$+1), where $X$ is an expoentially distributed random variable with parameter 1?

With parameter 1, $X$ can be written as $e^{-x}$, and after plug in the given function, I got $$\frac{1}{e^{-x}+1} = \frac{e^{x}}{e^{x}+1}$$
What type of distribution is this?

$X$ is not "writen" as $e^{-x}$.   The probability density function of $X$, called $f_X(x)$, is equal to $e^{-x}~\big[x\geqslant 0\big]$.
The cummulative distribution function of $X$ is: $$\begin{align}F_X(x) ~&=~ \mathsf P(X\leqslant x) \\[1ex] &=~ (1-e^{-x}) ~\big[x\geqslant 0\big]\end{align}$$
Let $Y:=1/(1+X)$.   The cummulative distribution function, and therefore probability density function, of $Y$ is
$$\begin{align}F_Y(y) ~&=~ \mathsf P(1/(1+X)\leqslant y) \\[1ex] &=~ \mathsf P(X\geqslant (1/y)-1) \\[1ex] &=~ 1- F_X(\tfrac 1y-1)\\[0ex] &~~\vdots\\[3ex] f_Y(y) ~&=~ \dfrac{\mathrm d ~F_Y(y)}{\mathrm d~y\qquad}\\[0ex] &~~\vdots\end{align}$$
