Arc length of $\alpha(t)=(t^6,t^9)$ I have to evaluate the arc length of the curve
$$\alpha(t)=(t^6,t^9)$$
for $t\in[-\sqrt[3]{2},\sqrt[3]{2}]$.
But when I do it, the final integral is
$$\int_{-\sqrt[3]{2}}^{\sqrt[3]{2}}3t^{5}\sqrt{4+9t^{6}}dt$$
which value is zero. What should I do?

Remember that $\sqrt{x^2}=|x|$.
So as mentioned by Arctic Char it should be $|t|^5$. Then you have $$\int_{-\sqrt[3]{2}}^{\sqrt[3]{2}}3|t|^{5}\sqrt{4+9t^{6}}dt=$$
$$=\int_{0}^{\sqrt[3]{2}}3t^{5}\sqrt{4+9t^{6}}dt-\int_{-\sqrt[3]{2}}^{0}3t^{5}\sqrt{4+9t^{6}}dt=$$
$$=\big[\frac{1}{27}(9t^6+4)^{\frac{3}{2}}\big]_{t=0}^{t=\sqrt[3]{2}}-\big[\frac{1}{27}(9t^6+4)^{\frac{3}{2}}\big]_{t=-\sqrt[3]{2}}^{t=0}$$
$$=\frac{1}{27}[40^{\frac{3}{2}}-8]-\frac{1}{27}[8-40^{\frac{3}{2}}]$$
$$=\frac{1}{27}[160\sqrt{10}-16]=\frac{16}{27}[10\sqrt{10}-1] \space(\approx18.14683)$$
(where $40^{\frac{3}{2}}=40\sqrt{40}=80\sqrt{10}$).

Or since the integrand is an even function you have $$\int_{-\sqrt[3]{2}}^{\sqrt[3]{2}}3|t|^{5}\sqrt{4+9t^{6}}dt$$
$$=2\int_{0}^{\sqrt[3]{2}}3t^{5}\sqrt{4+9t^{6}}dt$$
$$=2\cdot\frac{1}{27}[80\sqrt{10}-8]=\frac{16}{27}[10\sqrt{10}-1].$$
