Least Square Approximation, but cannot integrate this.

Let $V = \mathcal{C}[-1, 1]$ be the real vector space of real-valued continuous functions defined on the closed interval $[-1, 1]$. $V$ is an inner product space with the inner product $\langle f, g\rangle = \int_{-1}^{1} f(x)g(x) dx$.

Find the least square approximation to $p = x^{1/3}$ in $W = \textrm{span} \left\{q_o = 1, q_1=x, q_2 = x^2 - \frac{1}{3} \right \}$.

I thought about using $\textrm{proj}_W p$, but I ran into the trouble of evaluating $\int_{-1}^{1} x^{1/3} dx$.

Any ideas?

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What trouble? An antiderivative is $(3/4)x^{4/3}$. – André Nicolas Mar 20 '12 at 19:20
The domain of $x^{1/3}$ is x > 0 only. – Hawk Mar 20 '12 at 19:38
It is a matter of definition. Certainly $(-1/2)^{\pi}$ makes no sense. But often $x^{p/q}$ is thought of as defined even for negative $x$, if $p$ and $q$ are integers and $q$ is odd. – André Nicolas Mar 20 '12 at 19:42
wolframalpha.com/input/?i=Integrate[x^%281%2F3%29%2C[x%2C-1%2C1}] – Hawk Mar 20 '12 at 19:50
Well, as I wrote, it is a matter of definition. But the question, as partly quoted, seems to treat $x^{1/3}$ as a function continuous on $[-1,1]$. So the problem setter's definition seems in this case to coincide with mine. – André Nicolas Mar 20 '12 at 20:07

Since the question asks about the least square approximation to $p(x)=x^{1/3}$ using the norm of $V$, the function $p$ has to be an element of $V$: otherwise the question would not make sense. So we must assume that the formula $x^{1/3}$ is intended to define a continuous function on $[-1,1]$. It is widely understood that for odd integers $n$, $x\mapsto x^{1/n}$ is the inverse of $x\mapsto x^n$; both are bijections on $\mathbb R$.
Being odd, $p$ is orthogonal to even functions $q_0$ and $q_2$. Its projection onto $W$ is simply $\dfrac{(p,q_1)}{(q_1,q_1)} q_1$.