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?
 A: You've misunderstood the difference between a random variable $X$ and its probability density function $f_X(x)$.  These are not the same thing.  $X$ represents the value of the random outcome.  $f_X(x)$ represents a likelihood of observing a particular outcome.
With this in mind, given that $X \sim \operatorname{Exponential}(1)$, we have $$f_X(x) = e^{-x}, \quad x \ge 0,$$ and the cumulative distribution function $$F_X(x) = \Pr[X \le x] = 1 - e^{-x}, \quad x \ge 0.$$  Then let $Y = 1/(1+X)$, so that the CDF of $Y$ is $$F_Y(y) = \Pr[Y \le y] = \Pr[1/(1+X) \le y] = \Pr[1+X \ge 1/y] = \Pr[X \ge \tfrac{1}{y} - 1].$$  What we have done is express the CDF of $Y$ in terms of a probability on $X$ through a monotone variable transformation.  Then we use the CDF of $X$ to compute the resulting probability:  $$F_Y(y) = \Pr[X \ge \tfrac{1}{y} - 1] = 1 - \Pr[X \le \tfrac{1}{y} - 1] = 1 - (1 - e^{-(1/y - 1)}) = e^{1 - 1/y}.$$  What is the corresponding support of $Y$?  Well, we know $X \ge 0$.  So $1+X \ge 1$, which in turn implies $1/(1+X) \le 1$.  And since $X$ is always positive, so is $1/(1+X)$.  Therefore, $0 < Y \le 1$ and we would write the complete CDF as $$F_Y(y) = \begin{cases} 0, & y \le 0, \\
e^{1 - 1/y}, & 0 < y \le 1, \\
1, & y > 1.\end{cases}$$  The density would be written $$f_Y(y) = F'_Y(y) = \begin{cases} y^{-2} e^{1-1/y}, & 0 < y \le 1, \\ 0, & \text{otherwise}.\end{cases}$$
A: Keep it short and simple. Let $Y=\frac{1}{1+X}$ where $0<Y<1$. Then
$P(Y<y)=P\left(\frac{1}{1+X}<y\right)=P\left(\frac{1}{1+X}<y\right)=P(1<y+Xy)$
$=P\left(X>\frac{1}{y}-1\right)=e^{1-1/y} \ \  $ for $0<y<1$
To get the pdf differentiate. I think you can manage it.
A: $X$ 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
$$\begin{align}F_Y(y) ~&=~ \mathsf P(1/(1+X)\leqslant y) \\[1ex] &=~ \mathsf P(X\geqslant (1/y)-1) \\[1ex] &=~ 1- F_X(\tfrac 1y-1)\\[0ex] &~~\vdots\\[3ex] f_Y(y) ~&=~ \dfrac{\mathrm d ~F_Y(y)}{\mathrm d~y\qquad}\\[0ex] &~~\vdots\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\int_{0}^{\infty}\expo{-x}\delta\pars{y - {1 \over x + 1}}\,\dd x & =
\int_{0}^{\infty}\expo{-x}\,
{\delta\pars{x -\bracks{1/y - 1}} \over \ds{%
\verts{\partiald{\bracks{y - 1/\pars{x + 1}}}{x}}_{\ x\ =\ 1/y - 1}}}\,\dd x
\\[5mm] & =
{1 \over y^{2}}\,\exp\pars{-\,\pars{{1 \over y} - 1}}
\int_{0}^{\infty}\delta\pars{x -\bracks{{1 \over y} - 1}}\,\dd x
\\[5mm] & =
{1 \over y^{2}}\,\exp\pars{1 - {1 \over y}}\bracks{{1 \over y} - 1 > 0}
\\[5mm] & =
{1 \over y^{2}}\,\exp\pars{1 - {1 \over y}}\bracks{y\pars{1 - y} > 0}
\\[5mm] & =
\bbx{\bracks{0 < y < 1}\,{1 \over y^{2}}\,\exp\pars{1 - {1 \over y}}}
\end{align}
