<s>[INST] Find the length of the parametric curve $x(t)=5+6t^4, \quad y(t)=5+4t^6\ , \quad0 ≤ t ≤ 2$ Find the length of the following parametric curve.
$$x(t)=5+6t^4\ ,\quad y(t)=5+4t^6\ ,\qquad    0  ≤  t  ≤  2.$$
I used the formula
$$\int_0^2\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$$
And I found 
$$\frac23\cdot 17^{3/2}+4-\frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.
My steps:
$$\left(\frac{dx}{dt}\right) = 24\cdot t^3 $$
$$\left(\frac{dy}{dt}\right) = 24\cdot t^5 $$
$$\int_0^2\sqrt{\left(24\cdot t^3\right)^2+\left(24\cdot t^5\right)^2}dt$$
$$\int_0^2\sqrt{\left(576\cdot t^6\right)+\left(576\cdot t^10\right)}dt$$
$$\int_0^2\sqrt{\left(576\cdot t^6\right) \cdot \left(1+t^4\right)}dt$$
$$24+\int_0^2\sqrt{\left(t^6\right) \cdot \left(1+t^4\right)}dt$$
$$\frac23\cdot 17^{3/2}+4-\frac23$$
 [/INST] Line 4 should read $$\int_{t=0}^2 \sqrt{576 t^6 + 576 t^{10}} \, dt.$$  This is a typesetting error.
Line 5 is correct.
Line 6 should read $$24 \int_{t=0}^2 \sqrt{t^6 (1+t^4)} \, dt.$$  The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.
You do not demonstrate how to proceed from Line 6 to Line 7.  I would complete the computation as follows:
$$\begin{align*}
24 \int_{t=0}^2 \sqrt{t^6(1+t^4)} \, dt
&= 24 \int_{t=0}^2 t^3 \sqrt{1+t^4} \, dt \qquad (u = 1+t^4; \; du = 4t^3 \, dt) \\
&= 6 \int_{u=1}^{17} \sqrt{u} \, du \\
&= 6 \left[\frac{2u^{3/2}}{3} \right]_{u=0}^{17} \\
&= 4 (17^{3/2} - 1) \\
&= 68 \sqrt{17} - 4.
\end{align*}$$
</s>