Finding density of $U = \frac{X}{X + Y}$ for $X, \ Y $ ~ $\text{Exp}(\lambda)$ i.i.d 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?
 A: Hint:
Consider $U=X+Y$ and $V=\frac{X}{X+Y}$. $U$ and $V$ have joint distribution given by
$$f_X(X(u,v))f_Y(Y(u,v)) J_\Phi(u,v)$$
where $J_\Phi(u,v)$ is the Jacobian determinant of the transformation $\Phi(u,v)=(uv,u-uv)$, and $f_X$, $f_Y$ are the density functions of $X$ and $Y$ (exponentials in your case)
A: Let $U={\large{\frac{X}{X+Y}}}$.

As you noted,
$$
\frac{x}{x+y}\le u
\iff
y\ge \frac{x(1-u)}{u}
$$
Using basically the same approach as in your attempt (and with some help from your comments), $F_U(u)$ can be computed as follows . . .
\begin{align*}
F_U(u)
&=
\int_0^\infty
\int_{{\Large{\frac{x(1-u)}{u}}}}^\infty
f_X(x)\,f_Y(y)
\;dy
\;dx
\\[4pt]
&=
\int_0^\infty
f_X(x)
\left(
\int_{{\Large{\frac{x(1-u)}{u}}}}^\infty
f_Y(y)
\;dy
\right)
\;dx
\\[4pt]
&=
\int_0^\infty
f_X(x)
\,\left(1-F_Y\Bigl(\frac{x(1-u)}{u}\Bigr)\right)
\;dx
\\[4pt]
&=
\int_0^\infty
\left(
\left(\lambda e^{-\lambda x}\right)
\left(
e^
{
-\lambda
\left(
{\Large{\frac{x(1-u)}{u}}}
\right)
}
\right)
\right)
\;dx
\\[4pt]
\end{align*}
which evaluates to $u$, hence
\begin{align*}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!
F_U(u)\,
&=
\begin{cases}
0&\text{if}\;\,u\le 0\\[4pt]
u&\text{if}\;\,0 < u < 1\\[4pt]
1&\text{if}\;\,u\ge 1\\[4pt]
\end{cases}
\\[10pt]
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!
f_U(u)
&=
\begin{cases}
1&\text{if}\;\,0 < u < 1\\[4pt]
0&\text{otherwise}\\[4pt]
\end{cases}
\\[4pt]
\end{align*}
A: We want to translate cordinates from $X,Y$ to $X,U$ where  $U=X/(X+Y)$, this implies $Y=X(1/U-1)$.
So the Jacobian matrix, and its absolute determinant, are: $$\begin{align}\mathcal J(x,u)&=\dfrac{\partial\langle x, x(1/u-1)\rangle}{\partial\langle x,u\rangle}\\[1ex]&=\begin{bmatrix}\partial x/\partial x & \partial x/\partial u\\ \partial(x(1/u-1))/\partial x& \partial(x(1/u-1))/\partial u\end{bmatrix}\\[1ex]&=\begin{bmatrix}1 & 0\\ (1/u-1)& -x/u^2\end{bmatrix}\\[2ex]\lVert\mathcal J(x,u)\rVert&=\lvert x\rvert/u^2\end{align}$$
Now the support for the $X,U$ distribution is $\{\langle x,u\rangle: 0<x, 0<x(1/u-1)\}\\=\{\langle x,u\rangle: 0<x, 0<u<1\}$
Which means $x$ has a strictly positive support.
Thus the probability density function is evaluated as:
$$\begin{align}f_{\small X,U}(x,u) &= \lVert\mathcal J(x,u)\rVert f_{\small X,Y}(x, x(1/u-1))\\[1ex]&=\lambda^2~x~\mathrm e^{-\lambda x/u}/u^2\cdot \mathbf 1_{0<x, 0<u<1}\\[2ex]f_{\small U}(u) &=\dfrac{\lambda^2~\mathbf 1_{ 0<u<1}}{u^2\qquad}\int_0^\infty x\,\mathrm e^{-x\lambda/u}\mathrm d x\\[1ex]&=\mathbf 1_{0<u<1} \end{align}$$
Therefore $U$ has a standard continuous uniform distribution. $U\sim\mathcal U(0..1)$, so:
$$F_{\small U}(u)= u\,\mathbf 1_{0\leqslant u<1}+\mathbf 1_{1\leqslant u}$$
A: Hint: I use some standerd transformation,
$$U= \frac{X}{(X+Y)} = \frac{1}{1 + \frac{Y}{X}}= \frac{1}{1+V}$$
$\frac{2X}{\lambda},\frac{2Y}{\lambda}\sim \chi^2_{(2)}$ independently. Therefore $V=\frac{Y}{X}\sim F_{(2,2)}$. Hence, $U\sim\operatorname{Beta}(1,1)\equiv U(0,1)$.
Here pdf of $F_{(2,2)}$,
$$
f(v)=\frac{1}{\operatorname{Beta}(1,1)} (1+v)^{-2} = \frac{1}{(1+v)^2}
$$
