Solution 1:

All of them can be solved using the beta function technique. See (I), (II), (III). Notice that, the second integral can be written as,

$$ \int_{-\infty }^{+\infty }{\frac{{{x}^{2m}}}{1+{{x}^{2n}}}\text{d}x}=2\int_{0 }^{+\infty }{\frac{{{x}^{2m}}}{1+{{x}^{2n}}}\text{d}x}. $$