How to compute $\int_0^2(1+4t^2+9t^4)^{1/2}\text{d}t$? The original question was:
find the length $\ell$ of the curve $\gamma$ given the parametric equations:
$$x=t~~~~~ y=t^2~~~~~ z=t^3 $$ 
from $t=0$ to $t=2$
 A: I bet there must be some error in this question because the indefinite integral looks monstrous, as Kim Peek already forsaw. It must be kind of Elliptic one. (I calculated the integral with wolfram)

How ever numerically the value between $0$ and $2$ is $9.57057$ See here
A: For sure, this is a monster and numerical integration would be required.
However, I was able to get "reasonable" results expanding the integrand as a Taylor series around $t=1$ and integrating between the given bounds. Since the integrand is now $$\sqrt{9 t^4+4 t^2+1}=\sum_{i=0}^n \alpha_i(t-1)^i+O\left((t-1)^{n+1}\right)$$ $$\int_0^2\sqrt{9 t^4+4 t^2+1}\,dt=\sum_{i=0}^n \alpha_i\int_0^2(t-1)^i\,dt=\sum_{i=0}^n \alpha_i \frac{1+(-1)^i}{1+i}=2\sum_{i=0}^n  \frac{\alpha_{2i}}{2i+1}$$ 
I give below a table for a few results
$$\left(
\begin{array}{cc}
n & integral \\
 0 & 7.48331 \\
 2 & 9.57050 \\
 4 & 9.59621 \\
 6 & 9.58476 \\
 8 & 9.56654 \\
 10 & 9.56044 
\end{array}
\right)$$
which are very quickly "close" to the numerical value already reported in
loved.by.Jesus's answer.
