A Putnam Integral $\int_2^4 \frac{\sqrt{\ln(9-x)}\,dx}{\sqrt{\ln(9-x)} + \sqrt{\ln(x+3)}}.$ This is a Putnam Problem that I have been trying to solve (on and off) for two years, but I have failed. I am in Calculus BC. This problem comes from the book "Calculus Eighth Edition by Larson, Hostetler, and Edwards". This problem is at the end of the first section of the chapter 8 exercises. Here's the problem:
Evaluate $$\int_2^4 \frac{\sqrt{\ln(9-x)}\,dx}{\sqrt{\ln(9-x)} + \sqrt{\ln(x+3)}}.$$
Please. Any help is very much appreciated. So are solutions. Thank you!
Edit: I like the solution given, but I was interested to see if there is any other way of doing the problem? I'm excited to see the results.

Let
$$ \mathcal{I}=\int_{2}^{4}\dfrac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}}\,\mathrm{d}x $$
Now, use that
$$ \int_{a}^{b}f(x)\,\mathrm{d}x\overset{(1)}{=}\int_{a}^{b}f(a+b-x)\,\mathrm{d}x $$
Then,
$$ \mathcal{I}=\int_{2}^{4}\dfrac{\sqrt{\ln(3+x)}}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}}\,\mathrm{d}x $$
Add up these two integrals to get
$$ 2\mathcal{I}=\int_{2}^{4}\dfrac{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}}\,\mathrm{d}x $$
Thus,
$$ \mathcal{I}=1 $$
In order to prove $(1)$, write the integral using another variable, say, $t$:
$$ \int_{a}^{b}f(a+b-x)\,\mathrm{d}x=\int_{a}^{b}f(a+b-t)\,\mathrm{d}t $$
In the latter one, set $x=a+b-t$ and $\mathrm{d}t=-\mathrm{d}x$ and change the limits of integration to obtain
$$ \begin{aligned}
\int_{a}^{b}f(a+b-t)\,\mathrm{d}t&=-\int_{b}^{a}f(x)\,\mathrm{d}x\\
&=\int_{a}^{b}f(x)\,\mathrm{d}x.
\end{aligned} $$
