# 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.

• 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. – Teepeemm Oct 4 '14 at 19:08
• 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. – Saudman97 Oct 4 '14 at 20:26

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}