The value of the integral \(\displaystyle\int_{0}^{\infty}\frac{dx}{(1+x^a)(1+x^2)}\ (a>0)\) is
$\frac{π}{4}$
Let \(I(a)=\displaystyle\int_0^\infty \frac{dx}{(1+x^a)(1+x^2)}\). Substitute \(x=1/t\), so \(dx=-dt/t^2\), and the limits \(0\to\infty\) map to \(\infty\to 0\).
This gives \(I(a)=\displaystyle\int_0^\infty \frac{t^2\,dt/t^2}{(1+t^{-a})(1+t^{-2})}=\int_0^\infty\frac{t^{a}\,dt}{(1+t^{a})(1+t^{2})}\) after clearing the negative powers.
Renaming \(t\) back to \(x\): \(I(a)=\displaystyle\int_0^\infty \frac{x^{a}\,dx}{(1+x^{a})(1+x^{2})}\).
Adding this to the original integral: \(2I(a)=\displaystyle\int_0^\infty \frac{1+x^{a}}{(1+x^{a})(1+x^{2})}\,dx=\int_0^\infty \frac{dx}{1+x^{2}}=\left[\tan^{-1}x\right]_0^\infty=\dfrac{\pi}{2}\).
Therefore \(I(a)=\dfrac{\pi}{4}\), independent of the value of \(a\) (for any \(a>0\)).
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