Putnam 1987 : B1

[Tolaso J Kos]

Evaluate the integral: $$\mathcal{I}=\int_{2}^{4}\frac{\sqrt{\ln \left ( 9-x \right )}}{\sqrt{\ln \left ( 9-x \right )}+\sqrt{\ln \left ( 3+x \right )}}\,{\rm d}x$$

[jacks]

Given \(\displaystyle \int_{2}^{4}\frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\)

Let \(\displaystyle I = \int_{2}^{4}\frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} = \int_{2}^{4}\frac{\sqrt{\ln(9-(2+4-x))} dx}{\sqrt{\ln(9-(2+4-x))}+\sqrt{\ln((2+4-x)+3)}}\)

Using \(\displaystyle \int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx\)

\(\displaystyle I = \int_{2}^{4}\frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} = \int_{2}^{4}\frac{\sqrt{\ln(x+3)} dx}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}}\)

\(\displaystyle 2I = \int_{2}^{4}1.dx = 2\Leftrightarrow I = 1\)

So \(\displaystyle I = \int_{2}^{4} \frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} = 1\)