How to prove this integration equality? How to prove 
\begin{align*}
&\mathrel{\phantom{=}}6\int_0^1\mathrm{d}x\int_1^{\infty}\frac{x\left( x-1 \right)}{t}\exp \left( -\frac{t}{\sqrt{x\left( 1-x \right)}} \right) \,\mathrm{d}t\\
&=-\int_1^{\infty}e^{-2x}\frac{2x^2+1}{2x^4}\sqrt{x^2-1}\,\mathrm{d}x
\end{align*}
I have no idea how to deal with the LHS.Any help will be grateful.

We can write
\begin{align}
I&=6\int_0^1\mathrm{d}x\int_1^{\infty}\frac{x\left( x-1 \right)}{t}\exp \left( -\frac{t}{\sqrt{x\left( 1-x \right)}} \right) \,\mathrm{d}t\\
&=6\int_0^1 x\left( x-1 \right)\mathrm{d}x\int_{\tfrac{1}{\sqrt{x\left( 1-x \right)}}}^{\infty}\frac{e^{-u}}{u} \,du\\
&=-12\int_0^{1/2} x\left( 1-x \right)\,dx\int_{\tfrac{1}{\sqrt{x\left( 1-x \right)}}}^{\infty}\frac{e^{-u}}{u} \,du\\
\end{align}
We take advantage of the symmetry with respect to $x=1/2$ for the las expression.
Now, with $y=\tfrac{1}{\sqrt{x\left( 1-x \right)}}$, one has $1-2x=\sqrt{1-\frac{4}{y^2}}$ and
$dx=-\frac{2}{y^2\sqrt{y^2-4}}dy$, and thus
\begin{align}
I&=-24\int_2^\infty \frac{dy}{y^4\sqrt{y^2-4}}\int_{y}^{\infty}\frac{e^{-u}}{u} \,du\\
&=-\frac{3}{2}\int_1^\infty \frac{dz}{z^4\sqrt{z^2-1}}\int_{2z}^{\infty}\frac{e^{-u}}{u} \,du
\end{align}
As
\begin{equation}
\int \frac{1}{v^4\sqrt{v^2-1}}dv=\frac{1}{3}\frac{\left( 2v^2+1 \right)\sqrt{v^2-1}}{v^3}
\end{equation} 
the last integral can be calculated by parts:
\begin{equation}
I=-\frac{1}{2}\int_1^\infty \frac{\left( 2z^2+1 \right)\sqrt{z^2-1}}{z^4}e^{-2z}\,dz
\end{equation}
