不定积分的计算

单变量函数的积分

张瑞
中国科学技术大学数学科学学院

有理函数的积分

定义 1.
两个实系数多项式的商，称为有理函数，其一般形式为

\[R(x)=\frac{P(x)}{Q(x)}=\frac{a_0+a_1 x+\cdots+a_nx^n}{b_0+b_1x+\cdots+b_mx^m} \]

其中$m,n$为非负整数，$a_n\neq0$, $b_m\neq0$，$a_i$, $b_i$均为实数。 $n<m$时，称为有理真分式，否则，称$R(x)$为有理假分式。

任何一个有理真分式可以拆解一个有理真分式与一个多项式的和。

几个简单有理函数的积分

\[\begin{aligned} \displaystyle\int\dfrac{1}{1+x^2}dx=&\arctan x \\ \displaystyle\int\dfrac{x}{1+x^2}dx=&\int\dfrac{1}{1+x^2}\frac{1}2d(x^2)=\frac{1}2\ln(1+x^2) \\ \end{aligned} \]

当$b^2-4ac<0$时，

\[\begin{aligned} \int\dfrac{px+q}{ax^2+bx+c}dx=&\int\dfrac{p(x+d)+q-pd}{a((x+d)^2+e)}dx \\ =&\dfrac{q-pd}a\int\dfrac1{(x+d)^2+e}d(x+d) \\ &+\frac{p}{2a}\int\dfrac1{(x+d)^2+e}d((x+d)^2) \\ \end{aligned} \]

当$b^2-4ac\geq0$时，上面的积分是多少？

$\displaystyle\int\dfrac1{(x^2+a^2)^n}dx$由前面给出了递推关系计算。

\[\begin{aligned} \int\frac{px+q}{(ax^2+bx+c)^n}dx=&\int\dfrac{p(x+d)+q-pd}{a^n((x+d)^2+e)^n}dx \\ =&\frac{p}{2a^n}\int\frac{x+d}{((x+d)^2+e)^n}d(x+d) \\ &+\frac{q-pd}{2a^n}\int\frac{1}{((x+d)^2+e)^n}d(x+d) \\ \end{aligned} \]

例 1. $\displaystyle\int\dfrac{x^2}{1+x^2}dx$

例 2. $\displaystyle\int\dfrac{x^3}{1+x^2}dx$

例 3. $\displaystyle\int\dfrac{x-1}{x^2+2x+3}dx$

例 4. $\displaystyle\int\dfrac{x-1}{x^2+2x-3}dx$

$\displaystyle\int\dfrac{x^2}{1+x^2}dx$

\[\begin{aligned} \int\dfrac{x^2}{1+x^2}dx =&\int\dfrac{1+x^2-1}{1+x^2}dx \\ =&\int 1 dx-\int\dfrac{1}{1+x^2}dx \end{aligned} \]
\[\begin{aligned} \int\dfrac{x^3}{1+x^2}dx =&\int\dfrac{x(1+x^2)-x}{1+x^2}dx \\ =&\int x dx-\int\dfrac{x}{1+x^2}dx \end{aligned} \]
\[\begin{aligned} &\int\dfrac{x-1}{x^2+2x+3}dx\\ =&\int\dfrac{x+1}{x^2+2x+3}dx-\int\dfrac{2}{x^2+2x+3}dx \\ =&\int\dfrac{\frac12}{x^2+2x+3}d(x^2+2x+3)-2\int\dfrac{1}{(x+1)^2+2}d(x+1) \\ =&\frac{1}2\ln(x^2+2x+3)-\sqrt2\arctan\dfrac{x+1}{\sqrt 2}+c \end{aligned} \]

若$Q(x)=(x-a)^k(x^2+px+q)^m$，则

\[\begin{aligned} \frac{P(x)}{Q(x)}=& \dfrac{c_1}{x-a}+\dfrac{c_2}{(x-a)^2}+\cdots+\dfrac{c_k}{(x-a)^k} \\ &+\dfrac{M_1x+N_1}{x^2+px+q}+\dfrac{M_2x+N_2}{(x^2+px+q)^2}+\cdots+\dfrac{M_mx+N_m}{(x^2+px+q)^m} \end{aligned} \]

例 5. $\displaystyle\int\dfrac{x}{(x+1)(x+2)(x+3)}dx$

例 6. $\displaystyle\int\dfrac{1}{(x+1)(x+2)^2(x+3)^3}dx$

\[\int\dfrac{x}{(x+1)(x+2)(x+3)}dx \]

设

\[\dfrac{x}{(x+1)(x+2)^2(x+3)^3}=\dfrac{A}{x+1}+\dfrac{B}{x+2}+\dfrac{C}{x+3} \]

由

\[\begin{aligned} \dfrac{x}{(x+2)^2(x+3)^3}={A}+\dfrac{B(x+1)}{x+2}+\dfrac{C(x+1)}{x+3} \\ \end{aligned} \]

取$x=-1$，则有

\[\dfrac{-1}2=A+0+0 \]
\[\int\dfrac{1}{(x+1)(x+2)^2(x+3)^3}dx \]

设

\[\begin{aligned} \dfrac{1}{(x+1)(x+2)^2(x+3)^3} =\dfrac{A}{x+1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2} \\ +\dfrac{D}{x+3}+\dfrac{E}{(x+3)^2}+\dfrac{F}{(x+3)^3} \end{aligned} \]
  • 两边乘$(x+1)$，取$x=-1$，得$A=\dfrac18$
  • 两边乘$(x+2)^2$，取$x=-2$，得$C=-1$
  • 两边乘$(x+3)^3$，取$x=-3$，得$F=-\dfrac12$

将$A,C,F$代入，得到

\[\dfrac{-x^2+x+22}{8(x+2)(x+3)^2}=\dfrac{B}{x+2}+\dfrac{D}{x+3}+\dfrac{E}{(x+3)^2} \]
  • 两边乘$(x+2)$，取$x=-2$，得$B=2$
  • 两边乘$(x+3)^2$，取$x=-3$，得$E=-\dfrac54$

最后，可得$D=-\dfrac{17}8$

例 7. $\displaystyle\int\dfrac{1}{x^3+1}dx$

例 8. $\displaystyle\int\dfrac{1}{x^4+1}dx$

例 9. $\displaystyle\int\dfrac{1}{x^6+1}dx$

\[\begin{aligned} \int\dfrac{1}{x^3+1}dx=\int\dfrac1{(x+1)(x^2-x+1)}dx \\ =\int\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2-x+1}dx \end{aligned} \]
  • 乘$x+1$，取$x=-1$，得$A=\dfrac13$
  • 然后，可得$B=-\dfrac13, C=\dfrac23$
\[\begin{aligned} \int\dfrac{1}{x^4+1}dx=&\int\dfrac{1}{(x^2+1)^2-2x^2}dx \\ =&\int\dfrac{ax+b}{x^2+\sqrt2x+1}+\dfrac{cx+d}{x^2-\sqrt2x+1}dx \end{aligned} \]

$b=d=\frac{1}2$， $a=-c=\dfrac{\sqrt2}4$

\[\begin{aligned} \int\dfrac{1}{x^6+1}dx=\int\dfrac1{(x^2+1)((x^2+1)^2-3x^2)}dx \\ =\int\dfrac{1}{(x^2+1)(x^2+1-\sqrt3x)(x^2+1+\sqrt3x)}dx \end{aligned} \]

若$Q(x)=(x-a)^k(x^2+px+q)^{m-1}$，则

\[\begin{aligned} \dfrac{P(x)}{Q(x)}=&(\dfrac{P_1(x)}{Q_1(x)})'+\dfrac{P_2(x)}{Q_2(x)} \\ =&(\dfrac{P_1(x)}{(x-a)^{k-1}(x^2+px+q)^{m-1}})'+\dfrac{P_2(x)}{(x-a)(x^2+px+q)} \end{aligned} \]

例 10. $\displaystyle\int\dfrac{4x^5-1}{(x^5+x+1)^2}dx$

例 11. $\displaystyle\int\dfrac{2x^4-4x^3+24x^2-40x+20}{(x-1)(x^2-2x+2)^3}dx$

\[\int\dfrac{4x^5-1}{(x^5+x+1)^2}dx \]
\[\begin{aligned} \dfrac{4x^5-1}{(x^5+x+1)^2}=&(\dfrac{Ax^4+Bx^3+Cx^2+Dx+E}{x^5+x+1})' \\ &+\dfrac{A_1x^4+B_1x^3+C_1x^2+D_1x+E_1}{x^5+x+1} \\ =&(\dfrac{-x}{x^5+x+1})' \end{aligned} \]

无理函数

$ \displaystyle\int\frac1{\sqrt{a^2-x^2}}dx=\arcsin\frac{x}{a}+C $

$\displaystyle\int\sqrt{a^2-x^2}dx=\frac{a^2}2\arcsin\frac{x}a+\frac12x\sqrt{a^2-x^2}+C $

$ \displaystyle\int\frac1{\sqrt{A+x^2}}dx=\ln|x+\sqrt{x^2+A}|+C $

$\displaystyle\int\sqrt{A+x^2}dx=\frac{A}2\ln(x+\sqrt{x^2+A})+\frac12x\sqrt{x^2+A}+C $

例 12. $\displaystyle\int\dfrac{x-1}{\sqrt{x^2+2x+3}}dx$

例 13. $\displaystyle\int\dfrac{1}{1+\sqrt x+\sqrt{1+x}}dx$

\[\begin{aligned} &\int\dfrac{x-1}{\sqrt{x^2+2x+3}}dx \\ =&\int\dfrac{\frac12}{\sqrt{x^2+2x+3}}d({x^2+2x+3})-\int\dfrac2{\sqrt{x^2+2x+3}}dx \\ =&\sqrt{x^2+2x+3}-\int\dfrac2{\sqrt{(x+1)^2+2}}d(x+1) \\ =&\sqrt{x^2+2x+3}-\ln(x+1+{\sqrt{(x+1)^2+2}})+c \end{aligned} \]
\[\begin{aligned} \int\dfrac{t^2}{t^4+1}dt =&\frac{1}2\int\dfrac{t^2-1}{t^4+1}dt+\frac{1}2\int\dfrac{t^2+1}{t^4+1}dt \\ =&\frac{1}2\int\dfrac{1-\frac1{t^2}}{t^2+\frac1{t^2}}dt+ \frac{1}2\int\dfrac{1+\frac1{t^2}}{t^2+\frac1{t^2}}dt \\ =&\frac{1}2\int\dfrac{1}{(t-\frac1{t})^2+2}d(t-\dfrac1t) \\ & +\frac{1}2\int\dfrac{1}{(t+\frac1{t})^2-2}d(t+\dfrac1t) \\ \end{aligned} \]
\[\begin{aligned} &\int\dfrac{1}{1+\sqrt x+\sqrt{1+x}}dx \\ =&\int\dfrac{1+\sqrt x-\sqrt{1+x}}{(1+\sqrt x-\sqrt{1+x})(1+\sqrt x+\sqrt{1+x})}dx \\ =&\int\dfrac{1+\sqrt x-\sqrt{1+x}}{2\sqrt x}dx \\ =&\sqrt x+\frac{x}2-\frac{1}2\int\sqrt{\dfrac{1+x}x} dx \end{aligned} \]

令$t=\sqrt{\dfrac{1+x}x}$，$x=\dfrac1{t^2-1}$， $dx=-\dfrac{2t}{(t^2-1)^2}dt$

\[\begin{aligned} -\frac{1}2\int\sqrt{\dfrac{1+x}x} dx =&\int\dfrac{t^2}{(t^2-1)^2}dt \\ =&\int(\dfrac{-\frac12 t}{t^2-1})'+\dfrac{\frac12}{t^2-1}dt \end{aligned} \]

令$x=(\dfrac{u^2-1}{2u})^2$，可以把$\sqrt x, \sqrt{1+x}$同时有理化

\[\int\dfrac1{1+\dfrac{u^2-1}{2u}+\dfrac{u^2+1}{2u}}dx \]

例 14. $\displaystyle\int\dfrac{x^2}{\sqrt{x^2+x+1}}dx$

例 15. $\displaystyle\int\dfrac{\sqrt{x^2+2x+2}}{x}dx$

例 16. $\displaystyle\int\dfrac{x^3}{\sqrt{1+2x-x^2}}dx$

\[\int\dfrac{x^2}{\sqrt{x^2+x+1}}dx=\int\dfrac{x^2}{\sqrt{(x+\frac12)^2+\frac34}}dx \]

$x+\frac12=t$

\[\begin{aligned} =&\int\dfrac{(t-\frac12)^2}{\sqrt{t^2+\frac34}}dt =\int\dfrac{t^2+\frac34-t-\frac12}{\sqrt{t^2+\frac34}}dt \\ =&\int\sqrt{t^2+\frac34}dt-\frac12\int\dfrac{1}{\sqrt{t^2+\frac34}}dt^2 -\frac12\int\dfrac1{\sqrt{t^2+\frac34}}dt \end{aligned} \]
\[\begin{aligned} \int\dfrac{\sqrt{x^2+2x+2}}{x}dx =&\int\dfrac{x^2+2x+2}{x\sqrt{x^2+2x+2}}dx \\ =&\int\dfrac{x+1}{\sqrt{(x+1)^2+1}}dx+\int\dfrac1{\sqrt{x^2+2x+2}}dx \\ &+\int\dfrac2{x\sqrt{x^2+2x+2}}dx \end{aligned} \]
\[\int\dfrac2{x\sqrt{x^2+2x+2}}dx =-\sqrt2\int\dfrac1{\sqrt{\frac1{x^2}+\frac1x+\frac12}}d\frac1x \]
\[\begin{aligned} &\int\dfrac{x^3}{\sqrt{1+2x-x^2}}dx \\ =&\int\dfrac{x^3}{\sqrt{2-(x-1)^2}}dx =\int\dfrac{(t+1)^3}{\sqrt{2-t^2}}dx \\ =&\int\dfrac{-t(2-t^2)}{\sqrt{2-t^2}}+\dfrac{-3(2-t^2)}{\sqrt{2-t^2}}+\dfrac{5t+7}{\sqrt{2-t^2}}dt \\ =&-\int t{\sqrt{2-t^2}}dt-3\int {\sqrt{2-t^2}}dt \\ &+5\int\dfrac{t}{\sqrt{2-t^2}}dt+7\int\dfrac1{\sqrt{2-t^2}}dt \\ \end{aligned} \]
\[\begin{aligned} &\int\dfrac{x^3}{\sqrt{1+2x-x^2}}dx \\ =&\dfrac13(2-t^2)^{\frac32} -3(\dfrac12t{\sqrt{2-t^2}}+\arcsin\dfrac{t}{\sqrt2}) \\ &-5{\sqrt{2-t^2}}+7\arcsin\dfrac{t}{\sqrt 2}+c \end{aligned} \]

$\displaystyle\int\dfrac{x^3}{\sqrt{1+2x-x^2}}dx$

\[\dfrac{p_n(x)}y=(q_{n-1}y)'+\dfrac{\lambda}y \]

$q_{n-1}(x)$ 为次数 $p_n(x)$低一次的多项式

令$y={\sqrt{1+2x-x^2}}$则

\[\begin{aligned} \dfrac{x^3}y =&((Ax^2+Bx+C)y)'+\dfrac{\lambda}{y} \\ =&(2Ax+B)y+(Ax^2+Bx+C)\dfrac{2-2x}{2y}+\dfrac{\lambda}y \\ \end{aligned} \]
\[\begin{aligned} \dfrac{x^3}y =&\dfrac1y((2Ax+B)(1+2x-x^2) \\ &+(1-x)(Ax^2+Bx+C)+\lambda) \end{aligned} \]
\[\left\{ \begin{aligned} -3A=1 \\ 5A-2B=0 \\ 2A+3B-C=0 \\ B+C+\lambda=0 \end{aligned}\right. \]

可解得系数

\[\begin{aligned} \Rightarrow \begin{cases} A=-\frac13 \\ B=-\frac56 \\ C=-\frac{19}6 \\ \lambda=4 \end{cases} \end{aligned} \]

若$P_n(x)$为$n$次多项式，$y=\sqrt{ax^2+bx+c}$，则有

\[\int\dfrac{P_n(x)}y dx=Q_{n-1}(x)y+\lambda\int\dfrac1y dx \]

其中$Q_{n-1}(x)$为$n-1$次多项式，$\lambda$为常数

对于$\displaystyle\int R(x,(\dfrac{\alpha x+\beta}{\gamma x+\delta})^{r},(\dfrac{\alpha x+\beta}{\gamma x+\delta})^{s},\cdots)dx$，其中$r,s$为有理数。可以取$r,s$的公分母$m$，就可以化为$\sqrt[m]{\dfrac{\alpha x+\beta}{\gamma x+\delta}}$

  • 若$p$为整数；可设数$r,s$有公分母$N$，令$x=t^N$，就有$dx=Nt^{N-1}dt$，可以实现有理数。
  • 若$q$为整数，令$t=\sqrt[n]{(a+bz)}=(a+bx^s)^{\frac1n}$，可以化为$R(z,\sqrt[n]{a+bz})$型
  • 若$p+q$为整数，则可以化为 $=\dfrac1s\int(\dfrac{a+bz}{z})^pz^{p+q}dx$ 为$R(z,\sqrt[n]{\dfrac{a+bz}{z}})$型，可以令$t=\sqrt[n]{\dfrac{a+bz}{z}}$

目录

本节读完

例 17. Thanks

17.