Suppose $X, Y$ are independent exponentially distributed random variables with parameter $\lambda$, $U=X/Y$, $V=XY \Rightarrow x=\sqrt{uv}, y=\sqrt{v/u}$. By the change of variables formula, we have the joint density of $U,V$
$$f_{U,V}(u,v)=f_X(x)f_Y(y)|J(x,y)|=\frac{\lambda^2}{2u}\mathrm{exp}\{-\lambda(\sqrt{uv}+\sqrt{v/u})\}$$
where the Jacobian determinant of $(x,y)$ was computed to be $1/2u$. However, integrating this result over $v>0$ to get the marginal density of $U$, which I know from elsewhere to be $f_U(u)=\frac{1}{(u+1)^2}$, is giving me considerable difficulty. $U,V$ do not seem to be independent, so treating the $1/u$ term as constant isn't an option. Is there a way to transform back to $(x,y)$ and then integrate with respect to $xy$, or did I make a mistake calculating the joint distribution?
With the change of variables $x=uy$ we have $f_{U,Y}(u,y) = f_X(uy) \cdot f_Y(y) \cdot y$, so \begin{align} f_U(u) &= \int_0^\infty f_{U,Y}(u,y) \mathop{dy}\\ &= \int_0^\infty\lambda^2 y e^{-\lambda y(1+u)} \mathop{dy} \\ &= \frac{\lambda^2}{(\lambda(1+u))^2} \\ &= \frac{1}{(1+u)^2}. \end{align} Above, I used the fact $\int_0^\infty \frac{(\lambda(1+u))^2}{(2-1)!} y^{2-1} e^{-\lambda(1+u)y} \mathop{dy}=1$ by integrating the PDF of the $\text{Gamma}(2,\lambda(1+u))$ distribution.