Showing that $\int_{0}^{\infty} \frac{dx}{1 + x^2} = 2 \int_0^1 \frac{dx}{1 + x^2}$ I was reading an article in which it was stated that, with a change of variable, one could show that:
$$\int_{0}^{\infty} \frac{dx}{1 + x^2} = 2 \int_0^1 \frac{dx}{1 + x^2}$$
I tried with $t = 1 + \frac{1}{x}$ but that doesn't work out, especially because the lower bound doesn't become $0$.

With the substitution $u=\frac{1}{x}$ we get
$$\int_1^\infty \frac{dx}{1+x^2} = \int_0^1 \frac{u^{-2}}{1+u^{-2}}du=\int_0^1 \frac{du}{1+u^2}$$
which implies 
$$\int_0^\infty \frac{dx}{1+x^2} = \int_0^1 \frac{dx}{1+x^2} + \int_1^\infty \frac{dx}{1+x^2} = 2\int_0^1 \frac{dx}{1+x^2}$$
