How to prove ∫+∞−∞f(x)dx=∫+∞−∞f(x−1x)dx?\int_{-\infty}^{+\infty} f(x)dx = \int_{-\infty}^{+\infty} f\left(x – \frac{1}{x}\right)dx?

We can write
$$\begin{align} \int_{-\infty}^{\infty}f\left(x-x^{-1}\right)dx&=\int_{0}^{\infty}f\left(x-x^{-1}\right)dx+\int_{-\infty}^{0}f\left(x-x^{-1}\right)dx\\ &=\int_{-\infty}^{\infty}f(2\sinh\theta)\,e^{\theta}d\theta+\int_{-\infty}^{\infty}f(2\sinh\theta)\,e^{-\theta}d\theta\\ &=\int_{-\infty}^{\infty}f(2\sinh\theta)\,2\cosh\theta\,d\theta\\ &=\int_{-\infty}^{\infty}f(x)\,dx. \end{align}$$
To pass from the first to the second line, we make the change of variables $x=e^{\theta}$ in the first integral and $x=-e^{-\theta}$ in the second one.