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 ? 

$$ \begin{aligned}
\int\frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,\mathrm{d}x&=\int \frac{\sqrt{x^4 +1}}{x^3}\,\mathrm{d}x\\
&=\int\frac{x^4 + 1}{x^3\sqrt{x^4+1}}\,\mathrm{d}x\\
&=\int\frac{x}{\sqrt{\left(x^2\right)^2+1}}\,\mathrm{d}x + \int \frac{1}{x^5}\frac{1}{\sqrt{1 + 1/x^4}}\,\mathrm{d}x\\
&=\frac{1}{2}\ln\Big(x^2 + \sqrt{x^4 + 1}\Big) - \frac{1}{2}\sqrt{1+1/x^4} + C\\
&=\frac{1}{2}\ln\Big(x^2 + \sqrt{x^4 + 1}\Big) - \frac{\sqrt{x^4 + 1}}{2x^2}+C
\end{aligned} $$
