Q:

Finding arc length of $f(x) = \sqrt{x+2}$ - is there an easier way?

Find an arc length of $f(x) = \sqrt{x+2}$ $\quad \land \quad x\in<0,2>$
$$f'(x) = \frac{1}{2\sqrt{x+2}}$$
$$\int_{0}^{2} \sqrt{1+[f'(x)]^2}dx = $$
$$\int_{0}^{2} \sqrt{1+\Bigg[\frac{1}{2\sqrt{x+2}}\Bigg]^2}dx = $$
$$\int_{0}^{2} \sqrt{1+\frac{1}{4x+8}}dx = $$
$$\int_{0}^{2} \sqrt{\frac{4x+9}{4x+8}}dx = $$
$$\int_{0}^{2} \frac{4x+9}{\sqrt{(4x+8)(4x+9)}}dx = $$
$$\int_{0}^{2} \frac{4x+9}{\sqrt{16x^2 + 68x + 72}}dx = $$
$$\frac{1}{8} \int_{0}^{2} \frac{32x+68 + 4}{\sqrt{16x^2 + 68x + 72}}dx = $$
$$\frac{1}{8} \int_{0}^{2} \frac{32x+68}{\sqrt{16x^2 + 68x + 72}}dx + \frac{1}{2} \int_{0}^{2} \frac{dx}{\sqrt{16x^2 + 68x + 72}} = $$
$$\frac{1}{8} \int_{0}^{2} \frac{32x+68}{\sqrt{16x^2 + 68x + 72}}dx + \frac{1}{2} \int_{0}^{2} \frac{dx}{\sqrt{16x^2 + 68x + 72}} = $$
$$\frac{1}{8} \int_{72}^{272} \frac{dt}{\sqrt{t}} + \frac{1}{2} \int_{0}^{2} \frac{dx}{\sqrt{16\Big[ (x + \frac{17}{8})^2 + \frac{16}{1024}\Big]}} = $$
$$\frac{1}{8} \int_{72}^{272} \frac{dt}{\sqrt{t}} + \frac{1}{8} \int_{0}^{2} \frac{dx}{ (x + \frac{17}{8})^2 + \frac{1}{64}} = $$
$$\frac{1}{8} \int_{72}^{272} \frac{dt}{\sqrt{t}} + \frac{1}{8} \int_{\frac{17}{8}}^{\frac{33}{8}} \frac{du}{ u^2 + \frac{1}{64}} = $$
$$\frac{1}{4}\sqrt{t} \Bigg\rvert_{72}^{272}
 + \arctan{(8u)} \Bigg\rvert_{\frac{17}{18}}^{\frac{33}{8}} = $$
$$= \Bigg[ \frac{\sqrt{272} - \sqrt{72}}{4}\Bigg] \Bigg\rvert_{72}^{272} + \Bigg[ \arctan{33} - \arctan{17} \Bigg] \Bigg\rvert_{\frac{17}{18}}^{\frac{33}{8}} =
$$
$$= \Bigg[ \frac{2\sqrt{17} - 3\sqrt{2}}{2}\Bigg] \Bigg\rvert_{72}^{272} + \Bigg[ \arctan{33} - \arctan{17} \Bigg] \Bigg\rvert_{\frac{17}{18}}^{\frac{33}{8}}
$$
$$\approx  2.03$$
The result should be $\approx 2.08$.
So I got some mistake, not sure where.
Anyway, is there a friendlier way to find this arc length?
And also, can $\arctan{33} - \arctan{17}$ be further simplified?
Cheers

A:

You can use inverse function to calculate the same arc: $y=x^2-2$ is the inverse so the equivalent arc length will be
$$\int_{\sqrt{2}}^{2} \sqrt{1+4x^2}dx = \Bigg[\frac{1}{4}\ln(2x+\sqrt{1+4x^2})+\frac{1}{2}x\sqrt{1+4x^2}\Bigg] \Bigg\rvert_{\sqrt{2}}^{2}=\frac{1}{4}\Big(\ln (4+\sqrt{17})-\ln(2\sqrt{2}+3)\Big)+\sqrt{17}-1.5\cdot \sqrt{2} \approx 2.08 $$
(the integral can be solved by trig substitution $2x=\tan u$ but I just looked at integral tables).

