Calculate the length of a polar curve 
Calculate the length of the polar curve $$\theta (r)=\frac{1}{2}\left( r+\frac{1}{r}\right)$$ from r = 1 to r = 3.

I understand mostly how to get the length of a polar curve by:
$$\int_{a}^{b} \sqrt[]{(f(\theta ))^{2}+(f'(\theta ))^{2}} \ d\theta $$
But in this exercise i dont get how to do it. Maybe i need to write the function $\theta (r)$ in terms of $\theta$
Any ideas or hints? Thanks
 A: As OP states "polar curve", then we can imagine it as curve on polar plane, or curve on Cartesian plane in polar coordinates. So, we can do it in 2 ways:

*

*Considering it as usual parametrical representation of curve on plane $(\boldsymbol r, \boldsymbol\theta)$ and considering $r$ as parameter we have $$\left(\frac{ds}{dr}\right)^2 =\left(\frac{dr}{dr}\right)^2+\left(\frac{d \theta}{dr}\right)^2$$
So, length can be calculated as
$$\int\limits_{1}^{3}\sqrt{1+\theta'^2(r)}dr $$


*Considering on plane $(\boldsymbol x, \boldsymbol y)$ and taking polar representation of curve by $x =r \cos \theta, y = r \sin \theta$. Using in last formulas $\theta = \theta(r)$ we can calculate
$$\frac{dx}{dr} = \cos \theta - r \frac{d\theta}{dr} \sin \theta$$
and
$$\frac{dy}{dr} = \sin \theta + r \frac{d\theta}{dr} \cos \theta$$
Using obtained we have
$$\left(\frac{ds}{dr}\right)^2 =\left(\frac{dx}{dr}\right)^2+\left(\frac{dy}{dr}\right)^2 =\\
=\left(\ \cos \theta - r \frac{d\theta}{dr} \sin \theta \right)^2 + \left(\ \sin \theta + r \frac{d\theta}{dr} \cos \theta \right)^2 =\\
=1+ r^2\left(\frac{d\theta}{dr}\right)^2$$
and length can be calculate as integral
$$\int\limits_{1}^{3}\sqrt{1+r^2\theta'^2(r)}dr$$
A: Building on my comment, since $rd\theta/dr=\tfrac12(r-\tfrac1r)$, $\int ds=\int_1^3\tfrac12(r+\tfrac1r)dr=2+\tfrac12\ln3$.
