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I found this Integral Online:

$$\int_2^4 \frac{ \sqrt{\ln(9-x)} }{ \sqrt{\ln(9-x)}+\sqrt{\ln(x+3)} } dx$$

I realize this has been asked before. My question doesn't so much concern how to get, but is more about an elementary logical point. The idea is to split the integral and use the substitution $u = 6-x$ like so:

$$\int_2^3 \frac{ \sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} dx + \int_2^3 \frac{\sqrt{\ln(u+3)}}{\sqrt{\ln(u+3)}+\sqrt{\ln(9-u)}} dx$$

My question is, what allows me to "put" $x$ back into the second integral, so that I can add the two together?

On second thought, I do have a question about actually solving the problem. How the heck does one see the symmetry about $x=3$? Is the graph of the integrand symmetric about this line? Where is the symmetry "a priori." I am would like to develop such intuition. I realize this takes practice, but there must have been some observation that tips one of to the symmetry about $x=3$.

EDIT: I see that this has been voted to close for being a possible duplicate. I don't think this is the case, since my primary question is about this logically subtlety which, as far as I can see, doesn't seem addressed in the other stackexchange posts. My question is really a general question about "variable transformations" of integrals that just so happened to manifest itself in this problem (a question I have actually had for quite some time).

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    Your question about the symmetry is much easier than the question about the integral. Note that x = 3, all the logs have the same argument - $6$. – 2017-02-10
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    Remember that integrals don't care what you label the variable. – 2017-02-10
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    @SeanRoberson So, in other words, when I "substitute" back in $x$, I am NOT doing this by letting $x=u$, but rather the integral is invariant under the variable transform $u \mapsto x$? Is that the idea? Somehow this still doesn't feel right...It still isn't clear to me. – 2017-02-10
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    @user193319 I have seen this result being called Shakespeare's theorem. Anyway, the general idea is that variable name switching does not change anything. For example, $\int_0^1 x dx =\int_0^1 u du$. – 2017-02-10

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