I need help evaluating the following improper integral:
$$\int_{0}^{\infty}\frac{dx}{(x^2+1)(x^k+1)}$$
My attempt:
$$\int_{0}^{\infty}\frac{dx}{x^{2+k}+x^k+x^2+1}$$
WLOG, Let $k = 0$. Then we have:
$$\frac{1}{2}\int_{0}^{\infty}\frac{dx}{(x^2+1)}$$
This is equal to $\frac{\tan^{-1}(x)}{2}$ evaluated at infinity which is equal to $\frac{\pi}{4}$. Is this correct? Do I lose generality by letting $k = 0$?
Let $\alpha$ be any real number.
$$\begin{align} I{\left(\alpha\right)} &=\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(1+x^{2}\right)\left(1+x^{\alpha}\right)}\\ &=\int_{0}^{\infty}\frac{y^{\alpha}}{\left(1+y^{2}\right)\left(1+y^{\alpha}\right)}\,\mathrm{d}y;~~~\small{\left[x=y^{-1}\right]}\\ &=\int_{0}^{\infty}\frac{1}{\left(1+y^{2}\right)}\left[1-\frac{1}{1+y^{\alpha}}\right]\,\mathrm{d}y\\ &=\int_{0}^{\infty}\frac{\mathrm{d}y}{1+y^{2}}-\int_{0}^{\infty}\frac{\mathrm{d}y}{\left(1+y^{2}\right)\left(1+y^{\alpha}\right)}\\ &=\int_{0}^{\infty}\frac{\mathrm{d}y}{1+y^{2}}-I{\left(\alpha\right)},\\ \end{align}$$
$$\implies I{\left(\alpha\right)}=\frac12\int_{0}^{\infty}\frac{\mathrm{d}y}{1+y^{2}}.$$
That is to say, your integral $I{\left(\alpha\right)}$ is in fact independent of $\alpha$!