Arclength Of $r = 1/\theta$

The general form of the arc-length of a polar curve is given by $L = \int_a^b \sqrt{r^2 + \Big(\frac{dr}{d\theta}\Big)^2} \, d\theta ,$ we know $r = \frac{1}{\theta}$ , $b=2$ and $a=1$ . This information gives us $\begin{aligned} L = & \int_1^2 \sqrt{\frac{1}{\theta^2} + \frac{1}{\theta^4}} \,d\theta \\ = & \int_1^2 \frac{1}{\theta^2}\sqrt{\theta^2+1}\,d\theta \\ = & \int_1^2 \frac{1}{\sqrt{\theta^2+1}} \,d\theta - \frac{\sqrt{\theta^2+1}}{\theta} \Big|_1^2 \, \, \, \, \, \, \, \, \text{via integration by parts}\\ = & \arcsin\text{h}(\theta)\Big|_1^2 - \frac{\sqrt{\theta^2+1}}{\theta} \Big|_1^2 \\ = & \ln(\theta + \sqrt{\theta^2+1})\Big|_1^2 - \frac{\sqrt{\theta^2+1}}{\theta} \Big|_1^2 \\ = & \boxed{\frac{2\sqrt{2}-\sqrt{5}}{2} + \ln\Big(\frac{2+\sqrt{5}}{1+\sqrt{2}}\Big)} \end{aligned}$