A function $f\left(x\right)$ satisfies $f\left(x\right)=f\left(\frac{c}{x}\right)$ for some real number $c\left(c>1\right)$ and $\forall x>0$. If $\underset{1}{\overset{\sqrt{c}}{\int }}\frac{f\left(x\right)}{x}dx=3$, then the value of $\underset{1}{\overset{c}{\int }}\frac{f\left(x\right)}{x}dx$ is:

Let $u=\frac{c}{x}$, so $du=\frac{-c}{x^2}dx$,

So       $$\begin{gathered} \underset{1}{\overset{\sqrt{c}}{\int }}\frac{f\left(x\right)}{x}dx=\underset{1}{\overset{\sqrt{c}}{\int }}\frac{f\left(x\right)}{x^2}x dx \\ x=\frac{c}{u}\Rightarrow dx=-c \frac{du}{u^2} \\ =\underset{c}{\overset{\sqrt{c}}{\int }}\frac{f\left({\displaystyle \frac{c}{u}}\right)}{u}du=\underset{\sqrt{c}}{\overset{c}{\int }}\frac{f\left(u\right)}{u}du\left(\because f\left(x\right)=f\left(\frac{c}{x}\right)\right) \end{gathered}$$

Therefore, $\underset{1}{\overset{c}{\int }}\frac{f\left(x\right)}{x}dx=\underset{1}{\overset{\sqrt{c}}{\int }}\frac{f\left(x\right)}{x}dx+\underset{\sqrt{c}}{\overset{c}{\int }}\frac{f\left(u\right)}{u}du=3+3=6$