Integral from 0 to ∞ [x^a/(x^b+1)] dx

\begin{align*}
& I(a, b) = \int_0^\infty \dfrac{x^a}{x^b+1} dx =\dfrac{\beta(\dfrac{a+1}{b}, 1-\dfrac{a+1}{b})}{b} \\& \\&{\rm{Proof}}: \\& \\&\therefore u=x^b \Longrightarrow \int_0^\infty \dfrac{u^{\frac{a-b+1}{b}}}{b(u+1)} du \\& \\&\therefore \int_0^\infty \dfrac{u^{\frac{a-b+1}{b}}}{b(u+1)} du = \dfrac{\beta(\dfrac{a+1}{b}, 1-\dfrac{a+1}{b})}{b} \\& \\&\therefore \dfrac{\beta(\dfrac{a+1}{b}, 1-\dfrac{a+1}{b})}{b} =
\dfrac{\pi}{b} \dfrac{1}{\sin{(\frac{\pi(a+1)}{b}})} \\& \\&{\rm{Example}}: \\& \\& \therefore a=0, b=2 \Longrightarrow I(0, 2) = \int_0^\infty \dfrac{1}{x^2+1} dx \\& \\&= \dfrac{\pi}{2} \dfrac{1}{\sin{(\frac{\pi}{2}})}= \dfrac{\pi}{2} \\& \\&\end{align*}

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