integral ( 1/x, x, 0,1)
Is that $\displaystyle \int_0^1 \frac{1}{x} dx$?
If so, note it will not exist at the bound: $\displaystyle x=0$ because not only is the function not continuous at that point (cannot divide by zero), the resultant integral $\displaystyle \ln(x)$ cannot take values less than or equal to zero.
You can notice that $\displaystyle \int_0^1 \frac{1}{x} dx=\int_1^\infty \frac{1}{x} dx$.
So we have $\displaystyle \int_1^\infty \frac{1}{x} dx=$ limit when b tends to $\displaystyle \infty$ of $\displaystyle \int_1^b \frac{1}{x}dx =$ limit when b tends to $\displaystyle \infty$ of $\displaystyle ln(x)$ evaluated in $\displaystyle b$ minus $\displaystyle ln(x)$ evaluated in $\displaystyle 1$. Which is equal to limit when b tends to $\displaystyle \infty$ of $\displaystyle ln(x)$ evaluated in $\displaystyle b$, which is $\displaystyle \infty$, thus divergent.
Ok. I said $\displaystyle \int_0^1 \frac{1}{x} dx=\int_1^\infty \frac{1}{x} dx$, this implies that the graph of $\displaystyle f(x)=\frac{1}{x}$ is symmetric with respect to the graph of $\displaystyle y=x$. It means that if $\displaystyle f(x)=\frac{1}{x}$, then $\displaystyle f^{-1}(x)=f(x)$. In other words, f has its own bijection equals to itself.
Now to prove it, you just have to show that $\displaystyle f(f^{-1}(x))=x$.
$\displaystyle f(f^1(x))$ is equals to $\displaystyle \frac{1}{f^1(x)}=\frac{\frac{1}{1}}{\frac{1}{x}}=x$. It is demonstrated. If we take a precise example maybe you'll understand better. If $\displaystyle x=5$, then $\displaystyle f(5)=\frac{1}{5}$. That means that to get 5 with our function, we need to evaluated it in $\displaystyle \frac{1}{5}$, which works. Maybe you will understand even better seeing the graph. It's trivial that the area under the curve of $\displaystyle f(x)=\frac{1}{x}$ from $\displaystyle 0$ to $\displaystyle 1$ is equal to the area from 1 to $\displaystyle \infty$. To be rigorous, just say f is its own bijection (and show it as I did for you), that should be enough.