Compute the line integral of $f (x, y) = x^2 + y^2$ over the parabola $y = x^2$ from the point $x = 0$ to $x = 1$. I ended up with 
$$\int_1^0 t^2(f^2+1)(1+4t^2)^{\frac{1}{2}}\,dt.$$ I don't know if I'm right from there and I wouldn't even know how to solve it from there anyway. My lecturer didn't bother with giving answers so I don't know.

$\int_C f(x,y)\|dr\|$
$x = t\\y = t^2\\ \frac {dx}{dt} = 1\\\frac {dy}{dt} = 2t\\\|dr\| = \sqrt {(\frac {dx}{dt})^2+(\frac {dy}{dt})^2}\ dt = \sqrt {1+4t^2}\ dt$
$\int_0^1 (t^2 + t^4)\sqrt {1+4t^2} \ dt$
Do you want to use a trig substitution or hyperbolics?
$t = \frac 12 \tan \theta\\dt = \frac 12 \sec^2\theta \ d\theta$ 
$\int_0^{\frac{\pi}{4}} (\frac 14 \tan^2\theta + \frac 1{16} \tan^4\theta)\frac 12\sec^3\theta\ d\theta\\
\int_0^{\frac {\pi}{4}} \frac 18 \tan^2\theta\sec^3\theta + \frac 1{32} \tan^4\theta\sec^3\theta d\theta$
vs.
$t = \frac 12 \sinh u\ du\\ dt = \frac 12 \cosh u\ du$ 
$\int_0^{\sinh^{-1} 2} \frac 18 \sin^2 u\cosh^2 u + \frac 1{32} \sinh^4 u\cosh^2 u du\\
\int_0^{\sinh^{-1} 2} \frac 18 \left(\frac {e^u - e^{-u}}{2}\right)^2\left(\frac {e^u + e^{-u}}{2}\right)^2 + \frac 1{32} \left(\frac {e^u - e^{-u}}{2}\right)^4\left(\frac {e^u + e^{-u}}{2}\right)^2 du\\
$
I think the second one looks easier.
The coeficients are going to get messy though
