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.

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