$$\int_1^a f\left(x^2+\frac{a^2}{x^2}\right) \frac{1}{x} \mathrm{~d} x=\int_1^a f\left(x+\frac{a^2}{x}\right) \frac{1}{x} \mathrm{~d} x .$$
【答案】 18. 【参考证明】设 $x=\sqrt{t}$ ，则有
\begin{aligned} &\int_1^a f\left(x^2+\frac{a^2}{x^2}\right) \frac{1}{x} \mathrm{~d} x=\frac{1}{2} \int_1^{a^2} f\left(t+\frac{a^2}{t}\right) \frac{1}{t} \mathrm{~d} t \\ &=\frac{1}{2}\left[\int_1^a f\left(t+\frac{a^2}{t}\right) \frac{1}{t} \mathrm{~d} t+\int_a^{a^2} f\left(t+\frac{a^2}{t}\right) \frac{1}{t} \mathrm{~d} t\right] \end{aligned}

\begin{aligned} &\int_a^{a^2} f\left(t+\frac{a^2}{t}\right) \frac{1}{t} \mathrm{~d} t=-\int_a^1 f\left(u+\frac{a^2}{u}\right) \frac{1}{u} \mathrm{~d} u \\ &=\int_1^a f\left(u+\frac{a^2}{u}\right) \frac{1}{u} \mathrm{~d} u \end{aligned}

\begin{aligned} &\int_1^a f\left(x^2+\frac{a^2}{x^2}\right) \frac{1}{x} \mathrm{~d} x \\ &=\frac{1}{2}\left[\int_1^a f\left(t+\frac{a^2}{t}\right) \frac{1}{t} \mathrm{~d} t+\int_1^a f\left(u+\frac{a^2}{u}\right) \frac{1}{u} \mathrm{~d} u\right] \\ &=\int_1^a f\left(x+\frac{a^2}{x}\right) \frac{1}{x} \mathrm{~d} x . \end{aligned}