Aloha :)
Setze die Funktion in die Formel für die Bogenlänge \(\ell\) ein:$$\ell=\int\limits_{x=0}^1\sqrt{1+[f'(x)]^2}\,dx=\int\limits_{0}^1\sqrt{1+(2x)^2}\,dx$$
Dieses Integral ist sehr sehr schlimm. Daher machen wir einige Vorüberlegungen.
1) Ableitung der Tangens-Funktion:$$\left(\tan x\right)'=\left(\frac{\sin x}{\cos x}\right)'=\frac{\cos x\cdot \cos x-\sin x\cdot(-\sin x)}{\cos^2x}=\frac{\cos^2x+\sin^2x}{\cos^2x}=\left\{\begin{array}{l}1+\tan^2x\\[1ex]\frac{1}{\cos^2x}\end{array}\right.$$
2) Integral über den Kehrwert des Cosinus:$$\int\frac{1}{\cos x}\,dx=\int\frac{\cos x}{\cos^2 x}\,dx=\int\frac{\cos x}{1-\sin^2 x}\,dx$$Wir substituieren:$$u=\sin x\implies \frac{du}{dx}=\cos x\implies dx=\frac{du}{\cos x}\quad\leadsto$$$$\int\frac{1}{\cos x}\,dx=\int\frac{\cos x}{1-u^2}\cdot\frac{du}{\cos x}=\int\frac{du}{1-u^2}=\int\frac{du}{(1-u)(1+u)}$$$$\phantom{\int\frac{1}{\cos x}\,dx}=\frac12\int\frac{(1+u)+(1-u)}{(1+u)(1-u)}\,du=\frac12\int\left(\frac{1}{1-u}+\frac{1}{1+u}\right)\,du$$$$\phantom{\int\frac{1}{\cos x}\,dx}=\frac12\int\frac{1}{1-u}\,du+\frac12\int\frac{1}{1+u}\,du=-\frac12\ln|1-u|+\frac12\ln|1+u|$$$$\phantom{\int\frac{1}{\cos x}\,dx}=\frac12\ln\left|\frac{1+u}{1-u}\right|=\frac12\ln\left|\frac{1+\sin x}{1-\sin x}\right|$$
3) Bestimmung der Bogenlänge
Nach diesen kurzen Vorüberlegungen können wir das Integral berechnen. Wir substituieren:$$2x=\tan u\implies x=\frac12\tan u\implies\frac{dx}{du}\stackrel{(1)}{=}\frac12\left(1+\tan^2u\right)\implies dx=\frac12(1+\tan^2u)\,du$$Die Integrationsgrenzen werden zu:$$2x=\tan u\implies u(x)=\arctan(2x)\implies u(0)=\arctan(0)=0\;;\;u(1)=\arctan(2)$$
Für die Bogenlänge heißt das:$$\ell=\int\limits_{0}^1\sqrt{1+(2x)^2}\,dx=\int\limits_0^{\arctan(2)}\sqrt{1+\tan^2u}\cdot\frac12(1+\tan^2u)\,du$$
Wegen \(\left(1+\tan^2u=\frac{1}{\cos^2u}\right)\) bedeuet dies:$$\ell=\frac12\!\!\!\int\limits_0^{\arctan(2)}\!\!\!\underbrace{\frac{1}{\cos u}}_{=f}\cdot\underbrace{\frac{1}{\cos^2u}}_{=g'}\,du=\frac12\!\!\!\int\limits_0^{\arctan(2)}\!\!\!\underbrace{\frac{1}{\cos u}}_{=f}\cdot\underbrace{(1+\tan^2u)}_{=g'}\,du$$$$\phantom{\ell}=\frac12\left[\underbrace{\frac{1}{\cos u}}_{=f}\cdot\underbrace{\tan u}_{=g}\right]_0^{\arctan(2)}-\frac12\!\!\!\int\limits_0^{\arctan(2)}\!\!\!\underbrace{\frac{\sin u}{\cos^2u}}_{=f'}\cdot\underbrace{\tan u}_{=g}\,du$$$$\phantom{\ell}=\frac12\left[\sqrt{\frac{1}{\cos^2u}}\tan u\right]_0^{\arctan(2)}-\frac12\!\!\!\int\limits_0^{\arctan(2)}\!\!\!\frac{\sin^2u}{\cos^3u}\,du$$$$\phantom{\ell}=\frac12\left[\sqrt{1+\tan^2u}\,\tan u\right]_0^{\arctan(2)}-\frac12\!\!\!\int\limits_0^{\arctan(2)}\!\!\!\frac{1-\cos^2u}{\cos^3u}\,du$$$$\phantom{\ell}=\frac12\left(\sqrt{1+2^2}\cdot2\right)-\underbrace{\frac12\!\!\!\int\limits_0^{\arctan(2)}\!\!\!\frac{1}{\cos^3u}\,du}_{=\ell}+\frac12\!\!\!\int\limits_0^{\arctan(2)}\!\!\!\frac{\cos^2u}{\cos^3u}\,du$$$$\phantom{\ell}=\sqrt5-\ell+\frac12\!\!\!\int\limits_0^{\arctan(2)}\!\!\!\frac{1}{\cos u}\,du\stackrel{(2)}{=}\sqrt5-\ell+\frac12\left[\frac12\ln\left|\frac{1+\sin u}{1-\sin u}\right|\right]_0^{\arctan(2)}$$$$\phantom{\ell}=\sqrt5-\ell+\frac14\ln\left|\frac{1+\frac{2}{\sqrt5}}{1-\frac{2}{\sqrt5}}\right|=\sqrt5-\ell+\frac14\ln\left|\frac{\sqrt5+2}{\sqrt5-2}\right|$$
Wir fassen zusammen:$$\ell=\frac{\sqrt5}{2}+\frac18\ln\left(\frac{\sqrt5+2}{\sqrt5-2}\right)\approx1,47894286$$
