一类简单有理分式函数的积分解
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
