Show that the value of a definite integral is unity $$ \int_2^4\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log(3+x)}}dx=1$$

We show that $I=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}\,dx$ equals $1$, where $f(x)=\sqrt{\log(9-x)}$.
Proof. By making a substitution $y=6-x$, we get 
$$I=\int_4^2 -\frac{f(6-y)}{f(y)+f(6-y)}\,dy=\int_2^4 \frac{f(6-y)}{f(y)+f(6-y)}\,dy.$$
Therefore
$$\begin{align*}2I&=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}dx+\int_2^4 \frac{f(6-y)}{f(y)+f(6-y)}dy\\
&=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}dx+\int_2^4 \frac{f(6-x)}{f(x)+f(6-x)}dx=\int_2^4 1\,dx =2.\end{align*}$$
Edit. In the last part, $y$ is a dummy variable and can be changed to $x$ or any other variable you like.
Edit2. If you don't like the same $x$ being used, you could use another variable, say $s$,
so that with two substitutions
$$\begin{align*}2I&=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}dx+\int_2^4 \frac{f(6-y)}{f(y)+f(6-y)}dy\\
&=\int_2^4 \frac{f(s)}{f(s)+f(6-s)}ds+\int_2^4 \frac{f(6-s)}{f(s)+f(6-s)}ds=\int_2^4 1\,ds =2.\end{align*}$$
