Let $U_1 \sim U(0,1)$, $U_2 \sim U(0,1)$, $X_1 = U_1^{1/\alpha}$ and $X_2 = U_2^{1/\beta}$
$U_1$ is independent of $U_2$ and $X_1 +X_2 \leq 1$
Show that: $$\dfrac{X_1}{X_1 + X_2} \sim Beta(\alpha, \beta)$$
My attempt to solve this question was the following:
$$P\Bigg(\dfrac{X_1}{X_1 + X_2} < x\Bigg) = P\Bigg(\dfrac{X_1}{X_2} < \dfrac{x}{1-x}\Bigg)$$
$$P(X_1<x_1) = P(U^{1/\alpha} < x_1) = P(U < x_1^\alpha) = x_1^\alpha = F_{X_1}(x_1)$$
$$P(X_2<x_2) = P(U^{1/\beta} < x_2) = P(U < x_2^\beta) = x_2^\beta=F_{X_2}(x_2)$$
So, the joint distribution is the following:
$$f(x_1, x_2) = \alpha x_1^{\alpha-1}\beta x_2^{\beta-1}$$
And I proceeded to use the Jacobian method:
$$W = \dfrac{X_1}{X_2}; V = X_1+X_2$$
So $$X_1 = \dfrac{VW}{1+W}; X_2 = \dfrac{V}{1+W} $$
So the Jacobian matrix will be:
$$J=\begin{bmatrix} \dfrac{-v}{(1+w)^2} & \dfrac{1}{1+w}\\\\ \dfrac{v}{(1+w)^2} & \dfrac{w}{1+w} \end{bmatrix}$$
And the absolute value of the determinant is:
$$|\dfrac{-vw}{(1+w)^3} - \dfrac{v}{(1+w)^3}| = \dfrac{v}{(1+w)^2}$$
$$g(w,v) = \alpha\Big( \dfrac{vw}{1+w} \Big)^{\alpha-1} \beta \Big( \dfrac{v}{1+w} \Big)^{\beta-1} \dfrac{v}{(1+w)^2} $$
$$g(w) = \int_{0}^{1} g(w,v) dv = \dfrac{\alpha\beta w^{\alpha-1}}{(1+w)^{\alpha+\beta}} \int_{0}^{1} v^{\alpha +\beta-1} dv =\dfrac{\alpha\beta w^{\alpha-1}}{(1+w)^{\alpha+\beta}} \dfrac{1}{\alpha+\beta} $$
And now I think that I did something wrong, beacause when try to find the CDF I didn't get nothing similar to the beta CDF.