Problem:
Given $X, Y$ ~ $\text{Exp}(\lambda)$ i.i.d, find $f_U, \ F_U$ for $U := \frac{X}{X + Y}$.
My approach:
For a fixed $u > 0$, parameterize $\{ (x,y) | \frac{x}{x + y} = u \}$ = $\{ (x,y) | y = \frac{x (1 - u)}{u} \}$ = $\{ (x,\frac{x (1 - u)}{u}) | x \geq 0\}$ ($x \geq 0$ holds by $X$ ~ $\text{Exp}(\lambda)$).
Then, one can compute: $$\int_0^{+\infty}f_X(x) f_Y\left(\frac{x (1 - u)}{u}\right) \mathrm{d}x = \int_0^{+\infty} \lambda e^{-\lambda x} \lambda e^{-\lambda \frac{x (1 - u)}{u}} \mathrm{d}x = \lambda^2 \int_0^{+\infty} e^{-\lambda x \frac{1}{u}} \mathrm{d}x = \\ \lambda^2 \left(-\frac{u}{\lambda} e^{-\lambda x \frac{1}{u}} \biggr{\rvert}_0^{+\infty}\right) = \lambda^2 \left(\frac{u}{\lambda}\right) = \lambda u $$
My problem:
Given those computations, I arrived at the conclusion $f_U(u) = \lambda u$. Although Wolfram Alpha agrees with my computations, the master solution does not, as according to it $F_U (u) = u$ (and therefore $f_U = 1$).
It'd be great to get some help on where I went wrong. Given that Wolfram Alpha indicates correct computations, I believe my mistake to be conceptual.
On a general note: How would you rate my approach; are there better ways to tackle such problems?