Prove $\int_{0}^{\infty}{x^2\over (e^{x}-x^2)^2}dx=\sum_{n=1}^{\infty}{n(2n)!\over (n+1)^{2n+1}}$ $$I=\int_{0}^{\infty}{x^2\over (e^{x}-x^2)^2}dx=\sum_{n=1}^{\infty}{n(2n)!\over (n+1)^{2n+1}}$$
Applying partial fractions
$x^2=A(e^x-x^2)+B$
$$I=\int_{0}^{\infty}{e^x\over (e^x-x^2)^2}-{1\over e^x-x^2}dx$$
Let 
$$J=\int_{0}^{\infty}{1\over e^x-x^2}dx$$
Applying partial fractions
$1=A(e^x-x)+B(e^x+x)$
$$J={1\over 2}\int_{0}^{\infty}{e^{-x}\over e^x+x}+{e^{-x}\over e^x-x}dx={1\over 2}(A+B)$$ respectively.
$$\sum_{n=1}^{\infty}x^{n-1}e^{-nx}={1\over e^x-x}$$
Let us concentrate only on B for the moment
$${1\over 2}B={1\over 2}\sum_{n=1}^{\infty}\int_{0}^{\infty}x^{n-1}e^{-x(n+1)}dx={1\over 2}\sum_{n=2}^{\infty}\int_{0}^{\infty}x^{n-2}e^{-xn}dx$$
We can apply Laplace transform
$$B=\sum_{n=2}^{\infty}{(n-2)!\over n^{n-1}}$$
The method I am doing right now seem lengthy, so I don't think this is right approach to tackle this problem. Can someone show me another way to tackle this problem?

\begin{align}
 I = \int_0^\infty  {\frac{{x^2 }}{{\left( {e^x  + x^2 } \right)^2 }}dx}  = \int_0^\infty  {\frac{{x^2 e^{ - 2x} }}{{\left( {1 + x^2 e^{ -  x} } \right)^2 }}dx}  
\end{align}
Using the geometric series we have $$ \frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty  {x^n }  \Rightarrow \frac{1}{{\left( {1 - x} \right)^2 }} = \sum\limits_{n = 1}^\infty  {nx^{n - 1} }  $$
Substituting 
\begin{align}
 I = \int_0^\infty  {x^2 e^{ - 2x}  \cdot \sum\limits_{n = 1}^\infty  {(-1)^{n-1}n\left( {x^2 e^{ -  x} } \right)^{n - 1} } dx}  \\ 
  = \int_0^\infty  {\sum\limits_{n = 1}^\infty  {(-1)^{n-1}nx^{2n} e^{ -  (n+1)x} } dx}  \\ 
  = \sum\limits_{n = 1}^\infty  {(-1)^{n-1}n\int_0^\infty  {x^{2n} e^{ -  (n+1)x} dx} } \,\, \\ 
 \end{align}
which is the gamma function
