Find $ \int \sqrt{\frac{1}{\theta^2}+ \frac{1}{\theta^4}} d\theta$ Any Ideas ! is this integrable function
 A: $$I=\int\sqrt{\frac1{x^2}+\frac1{x^4}}dx=\int\frac{\sqrt{1+x^2}}{x^2}dx$$
Set $x=\tan y$ or $u=\dfrac x{\sqrt{1+x^2}}$
A: If you have known the formula of
$$ \int\frac{1}{\sqrt{1+t^2}} dt=\ln(t+\sqrt{1+t^2})+C,$$
you can try as follows (For convenience, I have set $t=\theta$):
\begin{align*}
\sqrt{\frac{1}{t^2}+\frac{1}{t^4}}dt&=\frac{1}{t^2}\cdot\sqrt{1+t^2}dt\\
&=-\sqrt{1+t^2}d\left(\frac{1}{t}\right)\\
&=d\left(\big(-\sqrt{1+t^2}\big)\cdot\frac{1}{t}\right)+\frac{1}{t}\cdot \frac{t}{\sqrt{1+t^2}}dt\\
&=d\left(-\frac{\sqrt{1+t^2}}{t}+\ln\Big(t+\sqrt{1+t^2}\Big)\right).
\end{align*}
As a result, we have that
$$\int\sqrt{\frac{1}{t^2}+\frac{1}{t^4}}dt=-\frac{\sqrt{1+t^2}}{t}+\ln\Big(t+\sqrt{1+t^2}\Big)+C.$$
A: 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,
$$
\int \frac{1}{\theta} \sqrt{ 1 + \frac{1}{\theta^2} } d \theta
= \int 1 - u^2 du = u - \frac{u^3}{3} + C 
= \frac{1}{3} \sqrt{1 + \frac{1}{\theta^2}} \left( 2 - \frac{1}{\theta^2} \right) + C
$$
