#Impropres Integrals
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first integral is equal to the second integral in b)
u = 1/t, du = -dt/t²
try this variable substitution from the integral in b)
$\int_{0}^{\infty} \frac{t^{\alpha}}{(1+t^2)(1+t^\alpha)}dt = \int_{0}^{\infty} \frac{1}{t^2 (1 + t^{-2})(1 + t^{-\alpha})} dt$
Rion
as i said, variable substitution can solve this problem