Theorem: Variance of Logistic Distribution

Let $X$ be a continuous random variable which satisfies the logistic distribution:
:$X \sim \map {\operatorname {Logistic} } {\mu, s}$
The variance of $X$ is given by:
:$\var X = \dfrac {s^2 \pi^2} 3$

Proof:
From the definition of the logistic distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {\map \exp {-\dfrac {\paren {x - \mu} } s} } {s \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2}$
From Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \int_{-\infty}^\infty x^2 \, \map {f_X} x \rd x - \paren {\expect X}^2$
So:
:$\ds \var X = \frac 1 s \int_{-\infty}^\infty \dfrac {x^2 \map \exp {-\dfrac {\paren {x - \mu} } s} } {\paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^2} \rd x - \mu^2$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^{-1}
      | c = Integration by Substitution
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = -\paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^{-2} \paren {-\frac 1 s \map \exp {-\dfrac {\paren {x - \mu} } s} }
      | c = Power Rule for Derivatives, Chain Rule for Derivatives and {{Corollary|Derivative of Exponential Function|1}}
}}
{{eqn | ll= \leadsto
      | l = \dfrac 1 u - 1
      | r = \paren {\map \exp {-\dfrac {\paren {x - \mu} } s} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \ln {\dfrac 1 u - 1}
      | r = -\dfrac {\paren {x - \mu} } s
      | c =
}}
{{eqn | ll= \leadsto
      | l = -s \map \ln {\dfrac {1 - u} u} + \mu
      | r = x
      | c =
}}
{{eqn | ll= \leadsto
      | l = \mu^2 -2 s \mu \map \ln {\dfrac {1 - u} u} + s^2 \map {\ln^2} {\dfrac {1 - u} u}
      | r = x^2
      | c =
}}
{{end-eqn}}
and also:
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \to -\infty} \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^{-1}
      | r = 0
      | c = 
}}
{{eqn | l = \lim_{x \mathop \to \infty} \paren {1 + \map \exp {-\dfrac {\paren {x - \mu} } s} }^{-1}
      | r = 1
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \var X
      | r = \int_{\to 0}^{\to 1}  \paren {\mu^2 - 2 s \mu \map \ln {\dfrac {1 - u} u} + s^2 \map {\ln^2} {\dfrac {1 - u} u} } \rd u - \mu^2
      | c = 
}}
{{eqn | r = \mu^2 \int_{\to 0}^{\to 1} \rd u - 2 s \mu \int_{\to 0}^{\to 1} \map \ln {\dfrac {1 - u} u} \rd u + s^2 \int_{\to 0}^{\to 1} \map {\ln^2} {\dfrac {1 - u} u} \rd u - \mu^2
      | c = 
}}
{{eqn | r = \mu^2 - 2 s \mu \paren {\int_{\to 0}^{\to 1} \map \ln {1 - u} \rd u - \int_{\to 0}^{\to 1} \map \ln u \rd u} + s^2 \paren {\int_{\to 0}^{\to 1} \map {\ln^2} {1 - u} \rd u - 2 \int_{\to 0}^{\to 1} \map \ln {1 - u} \map \ln u \rd u + \int_{\to 0}^{\to 1} \map {\ln^2} u \rd u} - \mu^2
      | c = Definite Integral of Constant and Difference of Logarithms
}}
{{eqn | r =  \mu^2 - 2 s \mu \paren {\paren {-1} - \paren {-1} } + s^2 \paren {2 - 2 \paren {2 - \dfrac {\pi^2} 6} + 2} - \mu^2
      | c = {{Lemma|Expectation of Logistic Distribution|1|proof = yes}}, {{Lemma|Expectation of Logistic Distribution|2|proof = yes}}, {{Lemma|Variance of Logistic Distribution|1}}, {{Lemma|Variance of Logistic Distribution|2}} and {{Lemma|Variance of Logistic Distribution|3}}
}}
{{eqn | r = \dfrac {s^2 \pi^2} 3
}}
{{end-eqn}}
{{qed}}