一类简单有理分式函数的积分解

对于刚开始学习微积分的同学，一定会遇到这样几道不定积分问题

$$\begin{align} \int \frac{1}{x+1}\, \mathrm{d}x \quad \ \int \frac{1}{x^2+1}\, \mathrm{d}x \quad \ \int \frac{1}{x^3+1}\, \mathrm{d}x \\ \int \frac{1}{x^4+1}\, \mathrm{d}x \quad \ \int \frac{1}{x^5+1}\, \mathrm{d}x \quad \ \int \frac{1}{x^6+1}\, \mathrm{d}x \end{align}$$

当分母次数为1或2时，答案是显然的

次数为3时，需要利用立方和公式将分母因式分解然后待定系数法裂项求解

次数为4时，可以利用组合积分法，将原式转换为求解以下两种不定积分的问题

$$\begin{align} I_1=\int \frac{x^2+1}{x^4+1}\, \mathrm{d}x \\ I_2=\int \frac{x^2-1}{x^4+1}\, \mathrm{d}x \end{align}$$

这两个不定积分只需要将分子分母同时除以 $x^2$ ，也很容易求得答案

次数为6时，可以利用倒代换，将原式改写为

$$\begin{align} \int \frac{1}{x^6+1}\, \mathrm{d}x &= \frac{1}{2}\cdot \int \frac{x^4+1}{x^6+1}\, \mathrm{d}x \\ &=\frac{1}{2}\cdot \int \frac{x^4+1-x^2}{x^6+1}\,\mathrm{d}x+\frac{1}{2}\cdot \int \frac{x^2}{x^6+1}\,\mathrm{d}x\\ &=\frac{1}{2}\cdot \int \frac{1}{x^2+1}\,\mathrm{d}x+\frac{1}{6}\cdot \int \frac{1}{x^6+1}\,\mathrm{d}\left(x^3\right)\\ &=\frac{1}{2}\cdot \arctan x+\frac{1}{6}\cdot \arctan x^3+C \end{align}$$

可以看出以上五个不定积分求解起来还是比较简单的，用到的都是大一所学的基础知识

那么，接下来让我们求解一下分母次数为5的情况

第一步还是因式分解

$$\int \frac{1}{x^5+1}\,\mathrm{d}x =\int \frac{1}{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}\,\mathrm{d}x$$

进一步分析四次项，记其为 ①，对 ① 式同除 $x^4$ 得到 ② 式

$$\begin{align} &x^4-x^3+x^2-x+1=0 \quad \left(①\right)\\ x^{-4}&-x^{-3}+x^{-2}-x^{-1}+1=0 \quad \left(②\right) \end{align}$$

则可以得出 ① 的根互为倒数，因为 ① 式与 ② 式是等价的

用待定系数法求对 ① 进一步因式分解

$$\begin{align} &\left(x^2+ax+1\right)\left(x^2+bx+1\right)=0\\ x^4+&\left(a+b\right)x^3+\left(ab+2\right)x^2+\left(a+b\right)x+1=0 \end{align}$$

对比系数可得

$$\begin{cases} a+b&=-1\\ \ \ \ ab &=-1 \end{cases}$$

得

$$\begin{cases} \ \ a&=\displaystyle \frac{\sqrt{5}-1}{2}\\ \\ \ \ b&=\displaystyle -\frac{\sqrt{5}+1}{2} \end{cases}$$

故原式化为

$$\begin{align} \int \frac{1}{x^5+1}\,\mathrm{d}x &=\int \frac{1}{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}\,\mathrm{d}x\\ &=\int \frac{1}{\left(x+1\right)\left(x^2+\frac{\left(\sqrt{5}-1\right)x}{2}+1\right)\left(x^2-\frac{\left(\sqrt{5}+1\right)x}{2}+1\right)}\,\mathrm{d}x\\ &=\int \frac{4}{\left(x+1\right)\left(2x^2+\left(\sqrt{5}-1\right)x+2\right)\left(2x^2-\left(\sqrt{5}+1\right)x+2\right)}\,\mathrm{d}x\\ &=\int \left[\frac{Ax+B}{2x^2+\left(\sqrt{5}-1\right)x+2}+ \frac{Cx+D}{2x^2-\left(\sqrt{5}+1\right)x+2}+\frac{E}{x+1}\right]\,\mathrm{d}x \end{align}$$

记

$$\begin{align} f\left(x\right)&=\frac{4}{\left(x+1\right)\left(2x^2+\left(\sqrt{5}-1\right)x+2\right)\left(2x^2-\left(\sqrt{5}+1\right)x+2\right)}\\ &=\frac{Ax+B}{2x^2+\left(\sqrt{5}-1\right)x+2}+ \frac{Cx+D}{2x^2-\left(\sqrt{5}+1\right)x+2}+\frac{E}{x+1}\\ \end{align}$$

分别求解系数

对于 $E$

$$\begin{align} E&=\lim_{\left(x+1\right)\rightarrow 0}f\left(x\right)\left(x+1\right)\\ &=\lim_{\left(x+1\right)\rightarrow 0}\frac{4}{\left(2x^2+\left(\sqrt{5}-1\right)x+2\right)\left(2x^2-\left(\sqrt{5}+1\right)x+2\right)}\\ &=\lim_{\left(x+1\right)\rightarrow 0} \frac{1}{x^4-x^3+x^2-x+1}\\ &=\frac{1}{5} \end{align}$$

同理，对于 $Ax+B$

$$\begin{align} Ax+B &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{4}{\left(x+1\right)\left(2x^2+\left(\sqrt{5}-1\right)x+2\right)}\\ &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{4}{\left(x+1\right)\left(\left(\sqrt{5}+1\right)x-2+\left(\sqrt{5}-1\right)x+2\right)}\\ &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{4}{2\sqrt{5}x\left(x+1\right)}\\ &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{4}{\sqrt{5}\left(2x^2+2x\right)}\\ &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{4}{\sqrt{5}\left(\left(\sqrt{5}+1\right)x-2+2x\right)}\\ &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{4}{\sqrt{5}\left(\left(\sqrt{5}+3\right)x-2\right)}\\ &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{4}{\sqrt{5}\left(\sqrt{5}+3\right)\left(x-\frac{2}{\sqrt{5}+3}\right)}\\ &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{4\left(\sqrt{5}+3\right)}{\sqrt{5}\left(\sqrt{5}+3\right)\left(3-\sqrt{5}\right)\left(x-\frac{2\left(3-\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\right)}\\ &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{4\left(\sqrt{5}+3\right)}{4\sqrt{5}\left(x-\frac{2\left(3-\sqrt{5}\right)}{4}\right)}\\ &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{2\left(3-\sqrt{5}\right)}{\sqrt{5}\left(2x-\left(3-\sqrt{5}\right)\right)}\\ &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{2\left(3-\sqrt{5}\right)\left(x+P\right)}{\sqrt{5}\left(2x-\left(3-\sqrt{5}\right)\right)\left(x+P\right)} \end{align}$$

令

$$\begin{align} \left(2x-\left(3-\sqrt{5}\right)\right)&\left(x+P\right)\\ &=2x^2-\left(3-\sqrt{5}\right)x+2Px-\left(3-\sqrt{5}\right)P\\ &=2x^2-\left(1+\sqrt{5}\right)x-\left(3-\sqrt{5}\right)P\\ \\ &\rightarrow P=1-\sqrt{5} \end{align}$$

故 $Ax+B$ 可进一步化简

$$\begin{align} Ax+B &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{2\left(3-\sqrt{5}\right)\left(x+1-\sqrt{5}\right)}{\sqrt{5}\left(2x-\left(3-\sqrt{5}\right)\right)\left(x+1-\sqrt{5}\right)}\\ &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{2\left(3-\sqrt{5}\right)\left(x+1-\sqrt{5}\right)}{\sqrt{5}\left(\left(2x^2-\left(\sqrt{5}+1\right)\right)x+\left(3-\sqrt{5}\right)\left(\sqrt{5}-1\right)\right)}\\ &=\lim_{\left(2x^2-\left(\sqrt{5}+1\right)x+2\right))\rightarrow 0}\frac{\left(\sqrt{5}+1\right)x-4}{-\sqrt{5}} \end{align}$$

得

$$\begin{cases} \ \ A &=\displaystyle -\frac{\sqrt{5}+1}{5}\\ \\ \ \ B &=\displaystyle \ \ \ \ \frac{4}{5} \end{cases}$$

同理，也可解得

$$\begin{cases} \ \ C &=\displaystyle \frac{\sqrt{5}-1}{5}\\ \\ \ \ D &=\displaystyle \ \ \ \ \frac{4}{5} \end{cases}$$

将 $A,B,C,D,E$ 代回原式整理得

$$\begin{align} \int \frac{1}{x^5+1}\,\mathrm{d}x &=\int \left[\frac{\left(\sqrt{5}-1\right)x+4}{5\left(2x^2+\left(\sqrt{5}-1\right)x+2\right)}- \frac{\left(\sqrt{5}+1\right)x-4}{2\left(2x^2-\left(\sqrt{5}+1\right)x+2\right)}+\frac{1}{5\left(x+1\right)}\right]\,\mathrm{d}x\\ &=\int \frac{\left(\sqrt{5}-1\right)x+4}{5\left(2x^2+\left(\sqrt{5}-1\right)x+2\right)}\,\mathrm{d}x-\int \frac{\left(\sqrt{5}+1\right)x-4}{2\left(2x^2-\left(\sqrt{5}+1\right)x+2\right)}\,\mathrm{d}x+\frac{1}{5}\ln |x+1|+C \end{align}$$

分别令

$$\begin{align} J&=\int \frac{\left(\sqrt{5}-1\right)x+4}{5\left(2x^2+\left(\sqrt{5}-1\right)x+2\right)}\,\mathrm{d}x\\ K&=\int \frac{\left(\sqrt{5}+1\right)x-4}{2\left(2x^2-\left(\sqrt{5}+1\right)x+2\right)}\,\mathrm{d}x \end{align}$$

对于 $J$

$$\begin{align} J&=\int \frac{\left(\sqrt{5}-1\right)x+4}{5\left(2x^2+\left(\sqrt{5}-1\right)x+2\right)}\,\mathrm{d}x\\ &=\frac{1}{20}\int \frac{\left(\sqrt{5}-1\right)\left(4x+\sqrt{5}-1\right)-\left(\sqrt{5}-1\right)^2+16}{2x^2+\left(\sqrt{5}-1\right)x+2}\,\mathrm{d}x\\ &=\frac{1}{20}\int \frac{\left(\sqrt{5}-1\right)d\left(2x^2+\left(\sqrt{5}-1\right)x+2\right)}{2x^2+\left(\sqrt{5}-1\right)x+2}\,\mathrm{d}x+\frac{1}{20}\int \frac{10+2\sqrt{5}}{2x^2+\left(\sqrt{5}-1\right)x+2}\,\mathrm{d}x\\ &=\frac{\left(\sqrt{5}-1\right)\ln |2x^2+\left(\sqrt{5}-1\right)x+2|+C}{20}+\frac{5+2\sqrt{5}}{10}\int \frac{\mathrm{d}x}{2x^2+\left(\sqrt{5}-1\right)x+2}\\ &=\frac{\left(\sqrt{5}-1\right)\ln |2x^2+\left(\sqrt{5}-1\right)x+2|+C}{20}+\frac{5+2\sqrt{5}}{10}\int \frac{\frac{1}{\sqrt{2}}\mathrm{d}\left(\sqrt{2}x+\frac{\sqrt{5}-1}{2\sqrt{2}}\right)}{\left(\sqrt{2}x+\frac{\sqrt{5}-1}{2\sqrt{2}}\right)^2+\frac{\sqrt{5}+5}{4}}\\ &=\frac{\left(\sqrt{5}-1\right)\ln |2x^2+\left(\sqrt{5}-1\right)x+2|}{20}+\frac{5+2\sqrt{5}}{10}\frac{\sqrt{2}\arctan \left(\frac{4x+\sqrt{5}-1}{\sqrt{2}\sqrt{\sqrt{5}}}+5\right)}{\sqrt{\sqrt{5}}+5}+C\\ &=\frac{\left(\sqrt{5}-1\right)\ln |2x^2+\left(\sqrt{5}-1\right)x+2|}{20}+\frac{\sqrt{10+2\sqrt{5}}\arctan \left(\frac{4x+\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)}{10}+C \end{align}$$

同理可求得 $K$

$$K=\frac{\left(\sqrt{5}+1\right)\ln |2x^2-\left(\sqrt{5}+1\right)x+2|}{20}-\frac{\sqrt{10-2\sqrt{5}}\arctan \left(\frac{4x-\sqrt{5}-1}{\sqrt{10-2\sqrt{5}}}\right)}{10}+C$$

故最终结果为

$$\begin{align} \int \frac{1}{x^5+1}\,\mathrm{d}x&=\frac{1}{5}\ln |x+1|+J-K+C\\ &=\frac{1}{5}\ln |x+1|+\frac{\left(\sqrt{5}-1\right)\ln |2x^2+\left(\sqrt{5}-1\right)x+2|}{20}+\frac{\sqrt{10+2\sqrt{5}}\arctan \left(\frac{4x+\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right)}{10}\\ &-\frac{\left(\sqrt{5}+1\right)\ln |2x^2-\left(\sqrt{5}+1\right)x+2|}{20}+\frac{\sqrt{10-2\sqrt{5}}\arctan \left(\frac{4x-\sqrt{5}-1}{\sqrt{10-2\sqrt{5}}}\right)}{10}+C \end{align}$$

其实没什么思维难度，只是硬算

但如果我们将这题推广到次数为 $n$ 的情况并改为求其在 $[0,+\infty)$ 上的定积分，那么我们会得到一个非常优美的结论

$$\int_{0}^{+\infty}\frac{1}{x^n+1}\,\mathrm{d}x$$

作换元，令 $\displaystyle u=\frac{1}{x^n+1}$，则有

$$\begin{align} \mathrm{d}x &=\left[-\frac{1}{n}\cdot \left(1-u\right)^{\frac{1}{n}-1}\cdot u^{-\frac{1}{n}}-\frac{1}{n}\cdot \left(1-u\right)^{\frac{1}{n}}\cdot u^{-\frac{1}{n}-1}\right]\cdot \mathrm{d}u\\ &=-\frac{1}{n}\cdot \left(1-u\right)^{\frac{1}{n}-1}\cdot u^{-\frac{1}{n}-1}\cdot \mathrm{d}u \end{align}$$

则原式变为

$$\begin{align} \int_{0}^{+\infty}\frac{1}{x^n+1}\,\mathrm{d}x&=\frac{1}{n}\cdot \int_{0}^{1}u^{1-\frac{1}{n}-1}\cdot \left(1-u\right)^{\frac{1}{n}-1}\cdot \mathrm{d}u\\ &=\frac{1}{n}\cdot \mathrm{B} \left(1-\frac{1}{n},\frac{1}{n}\right)\\ &=\frac{1}{n}\cdot \Gamma\left(1-\frac{1}{n}\right)\cdot \Gamma\left(\frac{1}{n}\right) \end{align}$$

根据余元公式

$$\Gamma\left(z\right)\cdot \Gamma\left(1-z\right)=\frac{\pi}{\sin\left(\pi z\right)}$$

则最终结果为

$$\int_{0}^{+\infty}\frac{1}{x^n+1}\,\mathrm{d}x=\frac{\displaystyle \frac{\pi}{n}}{\sin \displaystyle \frac{\pi}{n}}$$

当然，这个定积分也可以用留数法做，过程也挺简单的，这里就不写了

还有一种方法是利用倒代换结合 digamma 函数，请读者自行探索