cdf of $X/(X+Y)$, where $X$ and $Y$ are i.i.d. uniform

$P\left[R\leq r\right]=\int_{0}^{1}P\left[R\leq r\mid Y=y\right]dy=\int_{0}^{1}P\left[X\leq\frac{ry}{1-r}\right]dy$.

It is evident that $P\left[R\leq0\right]=0$ and $P\left[R\leq1\right]=1$ so we can focus on the $r$ in interval $\left(0,1\right)$.

For $\frac{r}{1-r}\leq1$ or equivalently $r\leq\frac{1}{2}$ this leads to $\frac{r}{1-r}\int_{0}^{1}ydy=\frac{r}{2-2r}$.

For $r>\frac{1}{2}$ this leads to $\frac{r}{1-r}\int_{0}^{\frac{1-r}{r}}ydy+\int_{\frac{1-r}{r}}^{1}dy=\frac{3}{2}-\frac{1}{2r}$.

EDIT

In a more general context:

If $X,Y$ are independent real-valued random variables and $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ is measurable and nonnegative then

$Ef\left(X,Y\right)=\int\int f\left(x,y\right)dF_{X}\left(x\right)dF_{Y}\left(y\right)$.

If you are looking for the CDF of $R=g\left(X,Y\right)$ then you can define $f$ by $\left(x,y\right)\mapsto1$ if $g\left(x,y\right)\leq z$ and $\left(x,y\right)\mapsto0$ otherwise.

This leads to $Ef\left(X,Y\right)=P\left[Z\leq z\right]$ and $P\left[Z\leq z\mid Y=y\right]=\int f\left(x,y\right)dF_{X}\left(x\right)$ and we find

$P\left[Z\leq z\right]=\int P\left[Z\leq z\mid Y=y\right]dF_{Y}\left(y\right)$.

This statement is even true if $X,Y$ are not independent, but that is another chapter. The method is convenient if $P\left[Z\leq z\mid Y=y\right]$ is easy to find.

OP wrote: Find the cdf of $R = \frac{X}{X+Y}$. How would you go to do that?

How would I go to do that? I would use a computer algebra system to do the nitty gritties for me. In this instance, $X$ and $Y$ have joint pdf $f(x,y)$:

_{(source: tri.org.au)}