Integral$\int_0^\infty \ln x\,\exp(-\frac{1+x^4}{2\alpha x^2}) \frac{x^4+\alpha x^2- 1}{x^4}dx$? I am trying to prove
$$
I:=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+\alpha x^2- 1}{x^4}dx=\frac{\sqrt{2\alpha^3 \pi}}{2\sqrt[\alpha]e},\qquad \alpha>0.
$$
Note: The proof below shows how this is just a Gaussian integral!
I am not sure how to start this one.  It seems very difficult to me However the answer is very nice.
I thought maybe trying to write $I(\alpha)$ and $I'(\alpha)$ to try and simplify things but it didn't help much.  at $x=0$ there seems to be a problem with the integrand also however I am not sure how to go about using this.  Perhaps we could try and use a series expansion for $e^x=\sum_{n=0}^\infty x^n / n!$ however the function $e^{-1/x^2}$ is well known that its taylor series is zero despite the function not being. The factor of $x^4+\alpha x^2-1$ has been giving me trouble with simplifying the integrand.  Thanks.
To those who just made an edit:  If you are looking for a +2, please edit something worthwhile.  I edited it back to what I had considering you didn't fix anything as is shown in the Edit History.
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$\ds{I\equiv\int_{0}^{\infty}\ln\pars{x}
     \exp\pars{-\,{1 + x^{4} \over 2\alpha x^{2}}}\,
     {x^4+\alpha x^2- 1 \over x^4}\,\dd x=
     {\root{2\alpha^{3}\pi} \over 2\root[\alpha]{\expo{}}}:\ {\large ?},
     \qquad \alpha > 0}$.

From @Chen Wang answer $\ds{\pars{~\mbox{line}\ 4~}}$:
  $$
I=\alpha\int_{0}^{\infty}{1 \over x^{2}}
\exp\pars{-\,{1 + x^{4} \over 2\alpha x^{2}}}\,\dd x
$$

With $\ds{\expo{\theta} = x}$:
\begin{align}
I&=\alpha\int_{-\infty}^{\infty}\expo{-2\theta}
\exp\pars{-\,{\cosh\pars{2\theta} \over \alpha}}\,\expo{\theta}\,\dd\theta
=2\alpha\int_{0}^{\infty}\cosh\pars{\theta}
\exp\pars{-\,{\cosh\pars{2\theta} \over \alpha}}\,\dd\theta
\end{align}

Since
  $\ds{\cosh\pars{2\theta} = \cosh^{2}\pars{\theta} + \sinh^{2}\pars{\theta}
= 2\sinh^{2}\pars{\theta} + 1}$ and
  $\ds{\totald{\sinh\pars{\theta}}{\theta} = \cos\pars{\theta}}$ we'll have:
  \begin{align}
\color{#44f}{\large I}&=
2\alpha\expo{-1/\alpha}\
\overbrace{\int_{0}^{\infty}\cosh\pars{\theta}
\exp\pars{-\,{2\sinh^{2}\pars{\theta} \over \alpha}}\,\dd\theta}
^{\ds{\mbox{Lets}\ u\ \equiv\ \sinh\pars{\theta}}}
={2\alpha \over \root[\alpha]{\expo{}}}\int_{0}^{\infty}\expo{-2u^{2}/\alpha}
\,\dd u
\\[3mm]&={2\alpha \over \root[\alpha]{\expo{}}}\,\root{\alpha \over 2}\
\underbrace{\int_{0}^{\infty}\expo{-u^{2}}\,\dd u}_{\ds{=\ {\root{\pi} \over 2}}}
=\color{#44f}{\large{\root{2\alpha^{3}\pi} \over 2\root[\alpha]{\expo{}}}}
\end{align}

A: $$\begin{align*}
I&=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+\alpha x^2- 1}{x^4}dx\\
&=\int_0^\infty \ln x\, d\left(-\alpha x^{-1}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right)\\
&=-\alpha\left(\left.\frac{\ln x}{x}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right|_0^\infty-\int_0^\infty \frac{1}{x}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) d\,\ln x\right)\\
&=\alpha\int_0^\infty \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) dx\\
&=\alpha\left(\int_0^1 \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\underbrace{\int_1^\infty \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx}_{x\to1/x}\right) \\
&=\alpha\left(\int_0^1 \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\int_1^0 -\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\right) \\
&=\alpha\int_0^1 (1+\frac{1}{x^2})\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\\
&=\alpha\int_0^1 \exp\left(-\frac{1}{\alpha}-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\
&=\alpha e^{-1/\alpha}\int_0^1 \exp\left(-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\
&=\alpha e^{-1/\alpha}\int_{-\infty}^0 \exp\left(-\frac{y^2}{2\alpha }\right)dy\\
&=\alpha e^{-1/\alpha}\sqrt{\frac{\alpha\pi}{2}}.
\end{align*}$$
