Find $ \int \sqrt{\frac{1}{\theta^2}+ \frac{1}{\theta^4}} d\theta$ Any Ideas ! is this integrable function

Just for fun, an alternate approach: If $x = 1 + \frac{1}{\theta^2}$, then $\frac{d x}{d \theta} = - \frac{2}{\theta^3} = \frac{1}{\theta} \frac{-2}{\theta^2}$. So, the integral of interest is transformed to 
$$
\int \frac{1}{\theta} \sqrt{ 1 + \frac{1}{\theta^2} } d \theta
= -2 \int (x-1) \sqrt{ x } d x
$$
Where $u = \sqrt{x}$, the integrand transforms to $\frac{(u^2 -1)u}{2u} du = \frac{1}{2} (u^2 - 1) \, du $. Hence,
$$
\int \frac{1}{\theta} \sqrt{ 1 + \frac{1}{\theta^2} } d \theta
= \int 1 - u^2 du = u - \frac{u^3}{3} + C 
= \frac{1}{3} \sqrt{1 + \frac{1}{\theta^2}} \left( 2 - \frac{1}{\theta^2} \right) + C
$$
