Answer by Fixed Point for Prove that integral is independant of its parameter

Add the two fractions together and you get

$$J_1+J_2 = \int_1^\infty {\left(\frac{1}{1+x^\alpha}+\frac{x^\alpha}{1+x^\alpha}\right) \frac {dx}{1+x^2}}=\int_1^{\infty}\frac{1+x^{\alpha}}{1+x^{\alpha}}\frac {dx}{1+x^2}=\int_1^{\infty}\frac {dx}{1+x^2}$$