Integration: Gauss quadrature formula of the Integral $\int_{-1}^1f(x)\sqrt{|x|}dx$

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I want to find the formula for the Gauss quadrature that integrate the Integral $\int_{-1}^1f(x)\sqrt{|x|}\, dx$ exactly for every cubic polynomial $f$.

What exactly do we have to do here?

Is $\sqrt{|x|}$ the weighting function?

Could you explain me all the procedures step by step?

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Since the integration rule is to be exact for polynomials of degree three, you will get four conditions that can set four parameters. So, we want a rule of the form $Q(f) = w_0 f(x_0) + w_1 f(x_1)$, such that $$ Q(1) = I(1), Q(x) = I(x), Q(x^2)=I(x^2), Q(x^3)=I(x^3). $$

This leads to the system $$ \begin{cases} w_0+w_1 = \frac 43\\ w_0 x_0 + w_1 x_1 = 0\\ w_0 x_0^2 + w_1 x_1^2=\frac 47\\ w_0x_0^3+w_1 x_1^3 = 0 \end{cases} $$

Now you just need to solve for $x_0, x_1, w_0, w_1$, which is not hard in this case, to get $$ Q/f)= \frac 23 f\left(-\sqrt{\frac 37}\right)+\frac 23 f\left(\sqrt{\frac 37}\right). $$