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?

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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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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}
