Differential Geometry-curves Let $c:[0,2] \to\Bbb R^3$ be the  curve given by $$c(t)=(\frac {t^3}{3},t^2,2t).$$ 
Then there exists an $m>0$ and a $C^{\infty}$ bijection $f:[0,m]\to [0,2]$ such that $f'(s)> 0$ for every $s\in[0,m]$ and  such that the curve $b:[0,m]\to \Bbb R^3$ given by $$b(s)=c(f(s))\qquad (*)$$ satisfies  $|b'(s)|=1$ for every $s\in[0,m]$.
The question is to find $m$.
A thought of mine is this one:
Let $f(t)=\int_{0}^{t} |c'(t)|dt$ for $t\in[0,m]$. Then $f'(t)>0$ and has all the required properties.
Thus $f(m)=2$ adn we have that $\int_{0}^{m} |c'(t)|dt=2\int_{0}^{m} \sqrt{t^4+t^2+1} dt=2$ or $\int_{0}^{m} \sqrt{t^4+t^2+1} dt=1$. From here i should find $m$. But the indefinite integral of this root function cannot be found and the definite integral of this it should be hard, and i also don't know how to find it either.
Any help? Can you find another way of solving the problem?
Thanks

You have a small error making the integral impossible. 
Go back to calculating $c'$:
$$
c'(t) = (t^2,2t,2)
$$
so
$$
|c'(t)| = \sqrt{t^4 + 4t^2 + 4} = \sqrt{(t^2 + 2)^2}
$$
