试求 $$\bex I=\int_2^4\frac{\sqrt{\ln (9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}\rd x. \eex$$
解答: $$\beex \bea I&=\int_4^2 \frac{\sqrt{\ln(t+3)}}{\sqrt{\ln(t+3)}+\sqrt{\ln(9-t)}}(-\rd t)\quad\sex{9-x=t+3}\\ &=\int_2^4 \frac{\sqrt{\ln (x+3)}}{\sqrt{\ln(9-x)}+\sqrt{(x+3)}}\rd x\equiv J\\ &=\frac{1}{2}(I+J) =\frac{1}{2}\int_2^4 \rd x=1. \eea \eeex$$
