设 $\alpha$ 为实数, 则 $\int_0^{+\infty} \frac{\mathrm{d} x}{\left(1+x^2\right)\left(1+x^a\right)}=$
【答案】 $ \frac{\pi}{4} $

【解析】 $$
\begin{aligned}
&\text { 【解析】 } \int_0^{+\infty} \frac{\mathrm{d} x}{\left(1+x^2\right)\left(1+x^\alpha\right)}=\int_0^1 \frac{\mathrm{d} x}{\left(1+x^2\right)\left(1+x^\alpha\right)}+\int_1^{+\infty} \frac{\mathrm{d} x}{\left(1+x^2\right)\left(1+x^\alpha\right)}=I_1+I_2 \text {, } \\
&I_1=\int_0^1 \frac{\mathrm{d} x}{\left(1+x^2\right)\left(1+x^a\right)} \stackrel{x=\frac{1}{t}}{=} \int_{+\infty}^1 \frac{1}{\left(1+\frac{1}{t^2}\right)\left(1+\frac{1}{t^a}\right)}\left(-\frac{1}{t^2}\right) \mathrm{d} t=\int_1^{+\infty} \frac{t^a}{\left(1+t^2\right)\left(1+t^a\right)} \mathrm{d} t, \\
&\text { 则 } \int_0^{+\infty} \frac{\mathrm{d} x}{\left(1+x^2\right)\left(1+x^\alpha\right)}=\int_1^{+\infty} \frac{x^\alpha}{\left(1+x^2\right)\left(1+x^\alpha\right)} \mathrm{d} x+\int_1^{+\infty} \frac{\mathrm{d} x}{\left(1+x^2\right)\left(1+x^\alpha\right)} \\
&=\int_1^{+\infty} \frac{1}{1+x^2} \mathrm{~d} x=\left.\arctan x\right|_1 ^{+\infty}=\frac{\pi}{4}
\end{aligned}
$$
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