收集与参数无关的有理广义积分

青青子衿

\begin{align*}
\int_0^{+\infty}\dfrac{1}{(1+x)^2\left(1+x^{\alpha}\right)}\mathrm{d}x&=\dfrac{1}{2}\\
\int_0^{+\infty}\dfrac{x}{(1+x)^4\left(1+x^{\alpha}\right)}\mathrm{d}x&=\dfrac{1}{12}\\
\int_0^{+\infty}\dfrac{1}{(1+x^2)\left(1+x^{\alpha}\right)}\mathrm{d}x&=\dfrac{\pi}{4}\\
\int_0^{+\infty}\dfrac{1}{(1+x+x^2)\left(1+x^{\alpha}\right)}\mathrm{d}x&=\dfrac{\pi}{3\sqrt{\,3}}\\
\int_0^{+\infty}\dfrac{1}{(1-x+x^2)\left(1+x^{\alpha}\right)}\mathrm{d}x&=\dfrac{2\pi}{3\sqrt{\,3}}\\
\int_0^{+\infty}\dfrac{1}{(1+\frac{1}{2}x+x^2)\left(1+x^{\alpha}\right)}\mathrm{d}x&=\dfrac{\pi+2\operatorname{arccot}\sqrt{15}}{\sqrt{15}}\\
\int_0^{+\infty}\dfrac{1}{(1-\frac{1}{2}x+x^2)\left(1+x^{\alpha}\right)}\mathrm{d}x&=\dfrac{\pi-2\operatorname{arccot}\sqrt{15}}{\sqrt{15}}\\
\end{align*}