Subsection 5.4.1 有理函数的积分
有理函数是指两个多项式 \(P(x)\) 与 \(Q(x)\) 的商, 即
\begin{equation}
R(x)=\frac{P(x)}{Q(x)}=\frac{a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n}}{b_{0} x^{m}+b_{1} x^{m-1}+\cdots+b_{m}}\tag{5.4.1}
\end{equation}
其中 \(m, n\) 是非负整数; \(a_{0}, a_{1}, \cdots, a_{n}, b_{0}, b_{1}, \cdots, b_{m}\) 是常数,且 \(a_{0} \neq 0, b_{0} \neq 0\text{.}\)
当 \(n \geqslant m\) 时, \(R(x)\) 称为有理假分式; 当 \(n<m\) 时, \(R(x)\) 称为有理真分式.
有理假分式 \(\frac{P(x)}{Q(x)}\) 总可以用 \(Q(x)\) 除 \(P(x)\) 化为一个多项式 \(T(x)\) 和一个真分式之和,而多项式的不定积分是容易求出的. 因此计算有理函数的不定积分主要是计算有理真分式的不定积分.
由于两个真分式的代数和仍是一个真分式, 因此更希望把一个真分式化为若干个简单的真分式之和.
关于有理真分式在代数学上有下面的结论:
设
(5.4.1)是有理真分式
\((n<m)\text{,}\) 不妨设
\(b_{0}=1\text{,}\) 则
\(Q(x)\) 总能分解成质因式之积, 即
\begin{equation}
Q(x)=(x-a)^{\alpha} \cdots(x-b)^{\beta}\left(x^{2}+p x+q\right)^{\mu} \cdots\left(x^{2}+r x+s\right)^{v} .\tag{5.4.2}
\end{equation}
\begin{equation*}
\begin{aligned}
\frac{P(x)}{Q(x)}= & \frac{A_{1}}{x-a}+\frac{A_{2}}{(x-a)^{2}}+\cdots+\frac{A_{\alpha}}{(x-a)^{\alpha}}+\cdots+\frac{B_{1}}{x-b}+\frac{B_{2}}{(x-b)^{2}}+\cdots+ \\
& \frac{B_{\beta}}{(x-b)^{\beta}}+\frac{M_{1} x+N_{1}}{x^{2}+p x+q}+\frac{M_{2} x+N_{2}}{\left(x^{2}+p x+q\right)^{2}}+\cdots+\frac{M_{\mu} x+N_{\mu}}{\left(x^{2}+p x+q\right)^{\mu}}+\cdots+ \\
& \frac{U_{1} x+V_{1}}{x^{2}+r x+s}+\frac{U_{2} x+V_{2}}{\left(x^{2}+r x+s\right)^{2}}+\cdots+\frac{U_{v} x+V_{v}}{\left(x^{2}+r x+x\right)^{v}},
\end{aligned}
\end{equation*}
其中 \(A_{1}, \cdots, A_{\alpha}, B_{1}, \cdots, B_{\beta}, M_{1}, \cdots, M_{\mu}, N_{1}, \cdots, N_{\mu}, U_{1}, \cdots, U_{v}, V_{1}, \cdots, V_{v}\) 都是常数.
由上述定理知, 任何有理真分式都能分解成如 \(\frac{A}{x-a}, \frac{B}{(x-a)^{k}}, \frac{C x+D}{x^{2}+p x+q}\text{,}\) \(\frac{E x+F}{\left(x^{2}+p x+q\right)^{k}}\) 的分式之和, 其中 \(A, B, C, D, E, F\) 为实数, 自然数 \(k>1\text{,}\) 方程 \(x^{2}+\) \(p x+q=0\) 没有实根, 即 \(p^{2}-4 q<0\text{,}\) 从而有理式的积分问题, 就归纳为下面四种简单分式的积分：
\begin{equation*}
\displaystyle \int \frac{A}{x-a} \mathrm{~d} x ; \displaystyle \int \frac{B}{(x-a)^{k}} \mathrm{~d} x ; \displaystyle \int \frac{C x+D}{x^{2}+p x+q} \mathrm{~d} x ; \displaystyle \int \frac{E x+F}{\left(x^{2}+p x+q\right)^{k}} \mathrm{~d} x \quad(k>1) .
\end{equation*}
现讨论上述四种函数类型的不定积分, 只要求出这些类型的不定积分, 有理式的积分问题就解决了.
(1)\(\displaystyle \int \frac{A}{x-a} \mathrm{~d} x=A \ln |x-a|+C\text{.}\)
(2)\(\displaystyle \int \frac{B}{(x-a)^{k}} \mathrm{~d} x=B \displaystyle \int \frac{1}{(x-a)^{k}} \mathrm{~d}(x-a)=\frac{B}{-k+1}(x-a)^{-k+1}+C\text{.}\)
(3)\(\displaystyle \int \frac{C x+D}{x^{2}+p x+q} \mathrm{~d} x \quad\left(p^{2}-4 q<0\right)\text{.}\)
\(\displaystyle \int \frac{C x+D}{x^{2}+p x+q} \mathrm{~d} x\)
\(=\displaystyle \int \frac{C\left(x+\frac{p}{2}\right)+D-\frac{1}{2} C p}{\left(x+\frac{p}{2}\right)^{2}+q-\frac{p^{2}}{4}} \mathrm{~d} x\)
\(=C \displaystyle \int \frac{x+\frac{p}{2}}{\left(x+\frac{p}{2}\right)^{2}+q-\frac{p^{2}}{4}} \mathrm{~d} x+\left(D-\frac{1}{2} C p\right) \displaystyle \int \frac{1}{\left(x+\frac{p}{2}\right)^{2}+q-\frac{p^{2}}{4}} \mathrm{~d} x\)
\(=\frac{C}{2} \ln \left|x^{2}+p x+q\right|+\frac{2 D-C p}{\sqrt{4 q-p^{2}}} \arctan \frac{2 x+p}{\sqrt{4 q-p^{2}}}+C\text{.}\)
(4) \(\displaystyle \int \frac{E x+F}{\left(x^{2}+p x+q\right)^{k}} \mathrm{~d} x \quad\left(p^{2}-4 q<0, k>1\right)\text{.}\)
令 \(t=x+\frac{p}{2}\text{,}\) 于是有
\begin{equation*}
\displaystyle \int \frac{E x+F}{\left(x^{2}+p x+q\right)^{k}} \mathrm{~d} x=\frac{E}{2} \displaystyle \int \frac{2 t}{\left(t^{2}+a^{2}\right)} \mathrm{d} t+\left(F-\frac{E p}{2}\right) \displaystyle \int \frac{1}{\left(t^{2}+a^{2}\right)^{k}} \mathrm{~d} t,
\end{equation*}
其中 \(a^{2}=q-\frac{p^{2}}{4}\text{,}\)
\begin{equation*}
\begin{aligned}
\displaystyle \int \frac{2 t}{\left(t^{2}+a^{2}\right)^{k}} \mathrm{~d} t & =\frac{1}{1-k} \cdot \frac{1}{\left(t^{2}+a^{2}\right)^{k-1}}+C \\
& =\frac{1}{1-k} \cdot \frac{1}{\left(x^{2}+p x+q\right)^{k-1}}+C .
\end{aligned}
\end{equation*}
下面举几个有理真分式的积分例子.
Example 5.4.1.
例 1 求不定积分 \(\displaystyle \int \frac{x}{\left(x^{2}+1\right)(x-1)} \mathrm{d} x\text{.}\)
Solution.
解 设 \(\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{A x+B}{x^{2}+1}+\frac{C}{x-1}\text{,}\) 比较等式两边的分子得
\begin{equation*}
x=(A x+B)(x-1)+C\left(x^{2}+1\right)=(A+C) x^{2}+(B-A) x+(C-B) .
\end{equation*}
再比较两端 \(x\) 的同次幂的系数有
\begin{equation*}
\left\{\begin{array}{l}
A+C=0, \\
-A+B=1, \\
-B+C=0
\end{array}\right.
\end{equation*}
解得 \(A=-\frac{1}{2}, B=\frac{1}{2}, C=\frac{1}{2}\text{,}\) 于是
\begin{equation*}
\begin{aligned}
\displaystyle \int \frac{x}{\left(x^{2}+1\right)(x-1)} \mathrm{d} x & =-\frac{1}{2} \displaystyle \int \frac{x-1}{x^{2}+1} \mathrm{~d} x+\frac{1}{2} \displaystyle \int \frac{1}{x-1} \mathrm{~d} x \\
& =-\frac{1}{4} \ln \left(1+x^{2}\right)+\frac{1}{2} \arctan x+\frac{1}{2} \ln |x-1|+C .
\end{aligned}
\end{equation*}
Example 5.4.2.
例 2 求不定积分 \(\displaystyle \int \frac{1}{x^{2}+2 x+3} \mathrm{~d} x\text{.}\)
Solution.
解 方程 \(x^{2}+2 x+3=0\) 无实根 \(\left(p^{2}-4 q=4-12=-8<0\right)\text{,}\) 将其配方成
\begin{equation*}
x^{2}+2 x+3=(x+1)^{2}+(\sqrt{2})^{2} \text {. }
\end{equation*}
故 \(\displaystyle \int \frac{1}{x^{2}+2 x+3} \mathrm{~d} x=\displaystyle \int \frac{1}{(x+1)^{2}+(\sqrt{2})^{2}} \mathrm{~d}(x+1)=\frac{1}{\sqrt{2}} \arctan \frac{x+1}{\sqrt{2}}+C\text{.}\)
Example 5.4.3.
例 3 求不定积分 \(\displaystyle \int \frac{3 x+5}{x^{2}+2 x+2} \mathrm{~d} x\text{.}\)
Solution.
解 \(\displaystyle \int \frac{3 x+5}{x^{2}+2 x+2} \mathrm{~d} x=\frac{3}{2} \displaystyle \int \frac{\mathrm{d}\left(x^{2}+2 x+2\right)}{x^{2}+2 x+2}+2 \displaystyle \int \frac{\mathrm{d}(x+1)}{(x+1)^{2}+1}\)
\begin{equation*}
=\frac{3}{2} \ln \left|x^{2}+2 x+2\right|+2 \arctan (x+1)+C .
\end{equation*}
Example 5.4.4.
例 4 求不定积分 \(\displaystyle \int \frac{x^{2}+1}{x(x-1)^{2}} \mathrm{~d} x\text{.}\)
Solution.
解 因为 \(\frac{x^{2}+1}{x(x-1)^{2}}=\frac{1}{x}+\frac{2}{(x-1)^{2}}\text{,}\) 所以
\begin{equation*}
\displaystyle \int \frac{x^{2}+1}{x(x-1)^{2}} \mathrm{~d} x=\displaystyle \int\left[\frac{1}{x}+\frac{2}{(x-1)^{2}}\right] \mathrm{d} x=\ln |x|-\frac{2}{x-1}+C .
\end{equation*}
Example 5.4.5.
例 5 求不定积分 \(\displaystyle \int \frac{x^{2}+2 x-1}{(x-1)^{2}\left(x^{2}-x+1\right)} \mathrm{d} x\text{.}\)
Solution.
解 因为 \(\frac{x^{2}+2 x-1}{(x-1)^{2}\left(x^{2}-x+1\right)}=\frac{2}{x-1}+\frac{2}{(x-1)^{2}}-\frac{2 x+1}{x^{2}-x+1}\text{,}\) 所以
\begin{equation*}
\begin{aligned}
\displaystyle \int \frac{x^{2}+2 x-1}{(x-1)^{2}\left(x^{2}-x+1\right)} \mathrm{d} x & =\displaystyle \int\left[\frac{2}{x-1}+\frac{2}{(x-1)^{2}}-\frac{2 x+1}{x^{2}-x+1}\right] \mathrm{d} x \\
& =\displaystyle \int \frac{2}{x-1} \mathrm{~d} x+\displaystyle \int \frac{2}{(x-1)^{2}} \mathrm{~d} x-\displaystyle \int \frac{2\left(x-\frac{1}{2}\right)+2}{\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}} \mathrm{~d} x \\
& =2 \ln |x-1|-\frac{2}{x-1}-\ln \left|x^{2}-x+1\right|-\frac{4}{\sqrt{3}} \arctan \frac{2 x-1}{\sqrt{3}}+C .
\end{aligned}
\end{equation*}
Example 5.4.6.
例 6 求不定积分 \(\displaystyle \int \frac{1}{x\left(x^{7}+2\right)} \mathrm{d} x\text{.}\)
Solution.
解 令 \(x=\frac{1}{t}, \mathrm{~d} x=-\frac{1}{t^{2}} \mathrm{~d} t\text{,}\) 于是
\begin{equation*}
\begin{aligned}
\displaystyle \int \frac{1}{x\left(x^{7}+2\right)} \mathrm{d} x & =\displaystyle \int \frac{t}{\left(\frac{1}{t}\right)^{7}+2}\left(-\frac{1}{t^{2}}\right) \mathrm{d} t=-\displaystyle \int \frac{t^{6}}{1+2 t^{7}} \mathrm{~d} t \\
& =-\frac{1}{14} \ln \left|1+2 t^{7}\right|+C=-\frac{1}{14} \ln \left|2+x^{7}\right|+\frac{1}{2} \ln |x|+C .
\end{aligned}
\end{equation*}
注意 代换 \(x=\frac{1}{t}\) 在有理分式函数中分母的阶数较高时常使用.
