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 ? 

$$I = \int \frac{\sqrt{x^4+1}}{x^3} \mathrm{d}x$$
Substitute $u=\sqrt{x^4+1},  \mathrm{d}u=\frac{2x^3}{\sqrt{x^4+1}}\mathrm{d}x$
$$\int \frac{u^2}{2(u^2-1)^{\frac{3}{2}}} \mathrm{d}u$$ 
Now substitute $u=\sec \theta$  $\mathrm{du} = \sec \theta \tan \theta \mathrm{d}\theta$
Then $(u^2-1)^{\frac{3}{2}} = \tan^3 \theta$
$$I = \frac{1}{2}\int \csc^2 \theta \sec \theta \mathrm{d}\theta = \frac{1}{2}\int (\cot^2 \theta+1) \sec \theta \mathrm{d}\theta$$
And take it from there
