Sorry...

Okay, here I've corrected my working:

Let $\displaystyle w_1=1, w_2=x, w_3 =x^2$ then

$\displaystyle v_1 = w_1 = 1$

$\displaystyle v_2 = x- \frac{w_2 . v_1}{\| v_1 \|^2}v_1$

$\displaystyle = x - \int^1_0 x dx= x-\frac{x^2}{2} = x- \frac{1}{2}$

$\displaystyle v_3 = x^2- (x - \frac{1}{2})- \frac{w_3.v_2}{\| v_2 \|^2} v_2$

$\displaystyle \frac{w_3.v_2}{\| v_2 \|^2} v_2 = \frac{x^2.(x-\frac{1}{2})}{\sqrt{(x^2 + \frac{1}{4}} )^2}=\frac{\int_0^1 x^3-\frac{x^2}{2}}{x^2+\frac{1}{4}}(x-\frac{1}{2})$ $\displaystyle = \frac{\frac{1}{12}}{x^2+ \frac{1}{4}}(x-\frac{1}{2})=\frac{2x-1}{24x^2-6}$

So $\displaystyle v_3 = x^2-(x-\frac{1}{2})- \frac{2x-1}{24x^2-6}$

Then we normalize them

$\displaystyle q_1 = \frac{v_1}{\|v_1\|}= \frac{1}{\sqrt{1}}$

$\displaystyle q_2 = \frac{v_2}{\|v_2\|}= \frac{x-\frac{1}{2}}{\sqrt{x^2-\frac{1}{4}}}$

This can't possibly be correct since the norm of a vector, just as the inner product of two vectors, must be a number...So: $\displaystyle \left|\left|x-\frac{1}{2}\right|\right|=\sqrt{\int^1_0\left(x-\frac{1}{2}\right)^2dx}$ $\displaystyle =\sqrt{\frac{1}{3}\left[\left(x-\frac{1}{2}\right)^3\right]^1_0}=\sqrt{\frac{1}{3}\left(\frac{1}{8}-\left(-\frac{1}{8}\right)\right)}=\sqrt{\frac{1}{12}}$ $\displaystyle =\frac{1}{2\sqrt{3}}$ . Now take it from here correcting the following. Tonio
$\displaystyle q_3 = \frac{v_3}{\|v_3\|}= \frac{x^2-(x-\frac{1}{2})- \frac{2x-1}{24x^2-6}}{\sqrt{(x^2-(x-\frac{1}{2})- \frac{2x-1}{24x^2-6})^2}}$

So, is this correct then? I'm not sure if I know how to check the conditions you mentioned that they are pair-wise orthogonal & of lenght 1...