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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.

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    $\begingroup$ In general, when there's a symmetry (such as around $x=3$ in this problem), it's a good idea to try to exploit that symmetry. Which is more easily said once you've seen the answer. $\endgroup$ – Teepeemm Oct 4 '14 at 19:08
  • $\begingroup$ Yes. I had a feeling that there was an inherent symmetry, but I was doubting myself so I didn't take a serious enough look at it. Glory is achieved by the fearless and not the second guessers. I need to start taking symmetry much more seriously! Thank you for the advice. $\endgroup$ – Saudman97 Oct 4 '14 at 20:26
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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 $$ \begin{aligned} \int_{a}^{b}f(a+b-t)\,\mathrm{d}t&=-\int_{b}^{a}f(x)\,\mathrm{d}x\\ &=\int_{a}^{b}f(x)\,\mathrm{d}x. \end{aligned} $$

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    $\begingroup$ There is an interesting (very) related article published in MATHEMATICS MAGAZINE VOL. 81, NO. 2, APRIL 2008, and entitled ''Lazy Student Integrals'' By GREGORY GALPERIN & GREGORY RONSSE. $\endgroup$ – Idris Dec 14 '14 at 5:10
  • $\begingroup$ Very Nice Solution. $\endgroup$ – juantheron Aug 12 '15 at 15:51

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