An orthogonal basis from the standard basis

Step 1. $\langle 1,1\rangle=\int_0^1 1\,dt=1$, so $u_1=1$.

Step 2. $\langle X,1\rangle=\int_0^1 t\,dt=1/2$. Orthogonal part:
$$w_2=X-\frac12.$$
Norm: $\|w_2\|^2=\int_0^1(t-1/2)^2\,dt=1/12$. So $u_2=2\sqrt3(X-1/2)=\sqrt3(2X-1)$.

Step 3. $\langle X^2,1\rangle=1/3$, $\langle X^2,u_2\rangle=\sqrt3\int_0^1 t^2(2t-1)\,dt=\sqrt3(1/2-1/3)=\sqrt3/6$. Orthogonal part:
$$w_3=X^2-\frac13-\frac{\sqrt3}{6}\cdot\sqrt3(2X-1)=X^2-\frac13-\frac12(2X-1)=X^2-X+\frac16.$$
Norm: $\|w_3\|^2=\int_0^1(X^2-X+1/6)^2\,dt=1/180$. So $u_3=6\sqrt5(X^2-X+1/6)$.

Conclusion. The orthonormal basis is $(1,\sqrt3(2X-1),6\sqrt5(X^2-X+1/6))$.