Finding $\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$ Finding $$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$$

I suppose I need integration by parts and trigo substitution
Let $u=\frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$
Let $dv = \sqrt{1+(\frac{1}{x^2})^2}$, $\frac{1}{x^2} = \tan{\theta}$. Is my substitution OK?
So $x=\frac{1}{\sqrt{\tan{\theta}}} \Rightarrow dx = -\frac{\sec^2{\theta}}{2\sqrt{\tan{\theta}}}\,d\theta$. But this will be very complicated to integrate later?
Am I supposed to be trying something else? 

UPDATE: An attempt
$$\int \frac{1}{x} \sqrt{1+\frac{1}{x^4}} dx$$
Let $u = \frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$
Let $dv = \sqrt{1+(\frac{1}{x^2})^2} dx$
Let $\alpha = \frac{1}{x^2} \Rightarrow d\alpha = -\frac{1}{2x^3}$
$dv=\sqrt{1+\alpha^2} d\alpha$
Let $\alpha = \tan{\theta} \Rightarrow d\alpha = \sec^2{\theta} d\theta$
$dv = \sqrt{1+\tan^2{\theta}} \sec^2{\theta} d\theta = \sec^3{\theta}$. Looks wrong here ? 

Usually we want to get rid of the square root. One way to do so is to bring in into the form $\sqrt{1+\cosh^2(u)}$ so lets try this in two steps
$$\frac{1}{x^2}=v \Rightarrow dv = \frac{-1}{2x^3}dx \Leftrightarrow dx = -2x^3 \, dv$$
$$ \int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx = \int \frac{-2x^3}{x}\sqrt{1+v^2}\,dv = -2\int \frac{1}{v}\sqrt{1+v^2}\,dv$$
With $v=\cosh(u) \Rightarrow dv = \sinh(u)\,du$ and $1+\cosh^2(u) = \sinh^2(u)$ follows
$$ ... = -2\int\frac{\sinh^2(u)}{\cosh(u)} \,du$$
This will become a bit messy from here on. Seeing how the final result is ...
$$ -\frac{1}{2} \sqrt{1+\frac{1}{x^4}}+\frac{1}{2}\sinh^{-1}\left(x^2\right) $$
