[高等数学] 有理函数的积分

一、知识点

两个多项式的商 $\frac{P(x)}{Q(x)}$ 称为有理函数，又称有理分式。

当分子多项式 $P (x)$ 的次数小于分母多项式 $Q (x)$ 的次数时，称这有理函数为真分式，否则称为假分式。

对于真分式 $\frac{P(x)}{Q(x)}$ ，如果分母可分解为两个多项式的乘积 $Q(x)=Q_1(x)Q_2(x)$ ，且 $Q_1(x)$ 与 $Q_2(x)$ 没有公因式，那么它可分拆成两个真分式之和 $\frac{P(x)}{Q(x)}=\frac{P_1(x)}{Q_1(x)}+\frac{P_2(x)}{Q_2(x)}$ , 该步骤称为把真分式化成部分分式之和。

如果 $Q_1(x)$ 或 $Q_2$ 还能再分解成两个没有公因式的多项式的乘积，就可再分拆成更简单的部分分式。

最后，有理函数的分解式中只出现多项式、 $\frac{P_1(x)}{(x-a)^k}$ 、 $\frac{P_2(x)}{(x^2+px+q)^l}$ 等三类函数（这里 $p^2-4q<0$ ， $P_1(x)$ 为小于 $k$ 次的多项式， $P_2(x)$ 为小于 $2 l$ 次的多项式）.针对各部分分别求积分。

二、练习题（求不定积分）

$\begin{aligned} 1. \int \frac{x^3}{x+3}dx &=\int \frac{x^3+27-27}{x+3}dx\\ &=\int \frac{(x+3)(x^2-3x+9)-27}{x+3}dx\\ &=\int (x^2-3x+9)dx-\int \frac{27}{x+3}dx\\ &=\frac{1}{3}x^3-\frac{3}{2}x^2+9x-27ln|x+3|+C \end{aligned}$

$\begin{aligned} 2. \int \frac{2x+3}{x^2+3x-10}dx &=\int \frac{d(x^2+3x-10)}{x^2+3x-10}\\ &=ln|x^2+3x-10|+C \end{aligned}$

$\begin{aligned} 3. \int \frac{x+1}{x^2-2x+5}dx &=\int \frac{x-1+2}{(x-1)^2+4}dx\\ &=\int \frac{x-1}{(x-1)^2+4}dx+\int \frac{2}{(x-1)^2+4}dx\\ &=\frac{1}{2} \int \frac{d[(x-1)^2+4]}{(x-1)^2+4}+\frac{1}{2}\int \frac{dx}{(\frac{x-1}{2})^2+1}\\ &=\frac{1}{2}ln(x^2-2x+5)+arctan\frac{x-1}{2}+C \end{aligned}$

$\begin{aligned} 4. \int \frac{dx}{x(x^2+1)} &=\int (\frac{1}{x}-\frac{x}{x^2+1})dx\\ &=ln|x|-\frac{1}{2}\int \frac{d(x^2+1)}{x^2+1}\\ &=ln|x|-\frac{1}{2}ln(x^2+1)+C \end{aligned}$

$\begin{aligned} 5. \int \frac{3}{x^3+1}dx &=\int (\frac{1}{x+1}-\frac{x-2}{x^2-x+1})dx\\ &=\int \frac{1}{x+1}dx-\frac{x-2}{x^2-x+1}dx\\ &=ln|x+1|-\int \frac{x-\frac{1}{2}}{x^2-x+1}dx+\frac{3}{2}\int \frac{dx}{x^2-x+1}\\ &=ln|x+1|-\frac{1}{2}\int \frac{2x-1}{x^2-x+1}dx+\frac{3}{2}\int\frac{dx}{(x-\frac{1}{2})^2+\frac{3}{4}}\\ &=ln|x+1|-\frac{1}{2}\int \frac{d(x^2-x+1)}{x^2-x+1}+\frac{3}{2}\int \frac{dx}{\frac{3}{4}[(\frac{2x}{\sqrt{3}}-\frac{1}{\sqrt{3}})^2+1]}\\ &=ln|x+1|-\frac{1}{2}ln(x^2-x+1)+\sqrt{3}\int \frac{d(\frac{2x}{\sqrt{3}}-\frac{1}{\sqrt{3}})}{(\frac{2x}{\sqrt{3}}-\frac{1}{\sqrt{3}})^2+1}\\ &=ln|x+1|-\frac{1}{2}ln(x^2-x+1)+\sqrt{3}arctan(\frac{2\sqrt{3}x}{3}-\frac{\sqrt{3}}{3})+C \end{aligned}$

$\begin{aligned} 6. \int \frac{x^2+1}{(x+1)^2(x-1)}dx\\ \end{aligned}$

设 $\frac{x^2+1}{(x+1)^2(x-1)}=\frac{Ax+B}{(x+1)^2}+\frac{C}{x-1}$ ，得 $A=\frac{1}{2}$ ， $B=-\frac{1}{2}$ ， $C=\frac{1}{2}$ ，则

$\begin{aligned} \int \frac{x^3+1}{(x+1)^2(x-1)}dx &=\int [\frac{x-1}{2(x+1)^2}+\frac{1}{2(x-1)}]dx\\ &=\frac{1}{2}\int [\frac{x+1}{(x+1)^2}-\frac{2}{(x+1)^2}+\frac{1}{x-1}]dx\\ &=\frac{1}{2}(ln|x+1|+\frac{2}{x+1}+ln|x-1|)+C\\ &=\frac{1}{x+1}+\frac{1}{2}ln|x^2-1|+C \end{aligned}$

$\begin{aligned} 7. &\int \frac{xdx}{(x+1)(x+2)(x+3)}\\ \end{aligned}$

设 $\frac{x}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}$ ，得 $A=-\frac{1}{2}$ ， $B = 2$ ， $C=-\frac{3}{2}$ ，则

$\begin{aligned} \int \frac{xdx}{(x+1)(x+2)(x+3)} &=\int [-\frac{1}{2(x+1)}+\frac{2}{x+2}-\frac{3}{2(x+3)}]dx\\ &=-\frac{1}{2}ln|x+1|+2ln|x+2|-\frac{3}{2}ln|x+3|+C &=\frac{1}{2}ln\frac{(x+2)^4}{|(x+1)(x+3)^3|}+C \end{aligned}$

$\begin{aligned} 8. \int \frac{x^5+x^4-8}{x^3-x}dx &=\int (\frac{x^5+x^4}{x^3-x}-\frac{8}{x^3-x})dx\\ &=\int \frac{x^4(x+1)}{x(x+1)(x-1)}dx-\int \frac{8}{x(x+1)(x-1)}dx\\ &=\int \frac{x^3}{x-1}dx-\int (\frac{4}{x+1}-\frac{8}{x}+\frac{4}{x-1})dx\\ &=\int \frac{x^3-1+1}{x-1}dx-4ln|x+1|+8ln|x|-4ln|x-1|\\ &=\int (x^2+x+1+\frac{1}{x-1})dx-4ln|x+1|+8ln|x|-4ln|x-1|\\ &=\frac{x^3}{3}+\frac{x^2}{2}+x+ln|x-1|-4ln|x+1|+8ln|x|-4ln|x-1|+C\\ &=\frac{x^3}{3}+\frac{x^2}{2}+x-3ln|x-1|-4ln|x+1|+8ln|x|+C \end{aligned}$

$\begin{aligned} 9. \int \frac{dx}{(x^2+1)(x^2+x)} &=\int [\frac{1}{x}-\frac{x+1}{2(x^2+1)}-\frac{1}{2(x+1)}]dx\\ &=ln|x|-\frac{1}{2} ln|x+1|-\frac{1}{4}\int (\frac{2x}{x^2+1}+\frac{2}{x^2+1})dx\\ &=ln|x|-\frac{1}{2}ln|x+1|-\frac{1}{4}ln(x^2+1)-\frac{1}{2}arctanx+C \end{aligned}$

$\begin{aligned} 10. \int \frac{1}{x^4-1}dx &=\int [\frac{1}{4(x-1)}-\frac{1}{2(x^2+1)}-\frac{1}{4(x+1)}]dx\\ &=\frac{1}{4}ln|x-1|-\frac{1}{2}arctanx-\frac{1}{4}ln|x+1|+C \end{aligned}$

$\begin{aligned} 11. \int \frac{dx}{(x^2+1)(x^2+x+1)} &=-\frac{1}{2}ln(x^2+1)+\frac{1}{2}\int \frac{d(x^2+x+1)}{x^2+x+1}+\frac{1}{2}\int \frac{dx}{(x+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}\\ &=\frac{1}{2}ln(x^2+x+1)-\frac{1}{2}ln(x^2+1)+\frac{\sqrt{3}}{3}arctan\frac{2x+1}{\sqrt{3}}+C \end{aligned}$

$\begin{aligned} 12. \int \frac{(x+1)^2}{(x^2+1)^2}dx &=\int \frac{x^2+1+2x}{(x^2+1)^2}dx\\ &=\int \frac{2x}{(x^2+1)^2}dx+\int \frac{1}{x^2+1}dx\\ &=-\frac{1}{1+x^2}+arctanx+C \end{aligned}$

$\begin{aligned} 13. \int \frac{-x^2-2}{(x^2+x+1)^2}dx &=-\int \frac{x^2+2}{(x^2+x+1)^2}dx\\ &=-\int \frac{x^2+x+1-x+1}{(x^2+x+1)^2}dx\\ &=-\int \frac{dx}{x^2+x+1}+\int \frac{x-1}{(x^2+x+1)^2}dx\quad (1) \end{aligned}$

以下分别计算 $(1)$ 式中的两个不定积分：

$\begin{aligned} (1-1): \int \frac{dx}{x^2+x+1} &=\int \frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}}\\ &=\frac{4}{3}\int \frac{dx}{(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})^2+1}\\ &=\frac{2\sqrt{3}}{3}\int \frac{d(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})}{(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})^2+1}\\ &=\frac{2\sqrt{3}}{3}arctan(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})+C_1 \end{aligned}$

$\begin{aligned} (1-2): \int \frac{x-1}{(x^2+x+1)^2} &=\frac{1}{2}\int \frac{2x+1-3}{(x^2+x+1)^2}dx\\ &=\frac{1}{2}\int \frac{2x+1}{(x^2+x+1)^2}dx-\frac{3}{2} \int \frac{dx}{(x^2+x+1)^2}\\ \end{aligned}$

以下分别计算 $(1 - 2)$ 中的两个不定积分：

$\begin{aligned} (1-2-1): \frac{1}{2}\int \frac{2x+1}{(x^2+x+1)^2}dx &=\frac{1}{2}\int \frac{d(x^2+x+1)}{(x^2+x+1)^2}\\ &=-\frac{1}{2(x^2+x+1)}+C_2 \end{aligned}$

$\begin{aligned} (1-2-2): \frac{3}{2}\int \frac{dx}{(x^2+x+1)^2}&=\frac{3}{2}\int \frac{dx}{[(x+\frac{1}{2})^2+\frac{3}{4}]^2}\\ &=\frac{4\sqrt{3}}{3}\int \frac{d(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})}{[(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})^2+1]^2}\\ \end{aligned}$

令 $tant=\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}}$ , 则:

$\begin{aligned} (1-2-2)&=\frac{4\sqrt{3}}{3}\int \frac{d(tant)}{(tan^2t+1)^2}\\ &=\frac{4\sqrt{3}}{3}\int \frac{sec^2tdt}{sec^4t}\\ &=\frac{4\sqrt{3}}{3}\int \frac{dt}{sec^2t}\\ &=\frac{4\sqrt{3}}{3}\int cos^2tdt\\ &=\frac{4\sqrt{3}}{3}\int \frac{cos2t+1}{2}dt\\ &=\frac{\sqrt{3}}{3}\int cos2t d2t+\frac{2\sqrt{3}}{3}t\\ &=\frac{\sqrt{3}}{3}sin2t+\frac{2\sqrt{3}}{3}t+C_3\\ &=\frac{2\sqrt{3}tant}{3(1+tan^2t)}+\frac{2\sqrt{3}}{3}t+C_3\\ &=\frac{2\sqrt{3}(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})}{3[1+(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})^2]}+\frac{2\sqrt{3}}{3}arctan(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})+C_3\\ &=\frac{2x+1}{2(x^2+x+1)}+\frac{2\sqrt{3}}{3}arctanx(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})+C_3 \end{aligned}$

$\begin{aligned} \therefore (1-2)&=(1-2-1)-(1-2-2)\\ &=-\frac{x+1}{x^2+x+1}-\frac{2\sqrt{3}}{3}arctan(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})+C_2+C_3\\ \end{aligned}$

$\begin{aligned} \therefore (1)&=-(1-1)+(1-2)\\ &=-\frac{2\sqrt{3}}{3}arctan(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})-\frac{x+1}{x^2+x+1}-\frac{2\sqrt{3}}{3}arctan(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})+C\\ &=-\frac{x+1}{x^2+x+1}-\frac{4\sqrt{3}}{3}arctan(\frac{2x}{\sqrt{3}}+\frac{1}{\sqrt{3}})+C \end{aligned}$

链接：本题用到的三角函数

$\begin{aligned} 14. \int \frac{dx}{3+sin^2}&=\int \frac{dx}{4-cos^2x}\\ &=\int \frac{dx}{cos^2x(\frac{4}{cos^2x}-1)}\\ &=\int \frac{sec^2dx}{4sec^2x-1}\\ &=\int \frac{d(tanx)}{4tan^2+3}\\ &=\frac{\sqrt{3}}{6}\int \frac{d(\frac{2}{\sqrt{3}}tanx)}{(\frac{2}{\sqrt{3}}tanx)^2+1}\\ &=\frac{\sqrt{3}}{6}arctan(\frac{2}{\sqrt{3}}tanx)+C \end{aligned}$

$\begin{aligned} 15. \int \frac{dx}{3+cosx}&=\int \frac{dx}{2+2cos^2\frac{x}{2}}\\ &=\int \frac{d(\frac{x}{2})}{1+cos^2\frac{x}{2}}\\ &=\int \frac{sec^2\frac{x}{2}d(\frac{x}{2})}{sec^2\frac{x}{2}+1}\\ &=\int \frac{d(tan\frac{x}{2})}{tan^2\frac{x}{2}+2}\\ &=\frac{1}{2}\int \frac{d(tan\frac{x}{2})}{(\frac{1}{\sqrt{2}}tan\frac{x}{2})^2+1}\\ &=\frac{1}{\sqrt{2}} \int \frac{d(\frac{1}{\sqrt{2}}tan\frac{x}{2})}{(\frac{1}{\sqrt{2}}tan\frac{x}{2})^2+1}\\ &=\frac{\sqrt{2}}{2}arctan(\frac{\sqrt{2}}{2}tan\frac{x}{2})+C \end{aligned}$

$\begin{aligned} 16. \int \frac{dx}{2+sinx}&=\int \frac{dx}{2+2sin\frac{x}{2}cos{x}{2}}\\ &=\int \frac{sec^2\frac{x}{2}d(\frac{x}{2})}{sec^2\frac{x}{2}+tan\frac{x}{2}}\\ &=\int \frac{d(tan\frac{x}{2})}{tan^2\frac{x}{2}+1+tan\frac{x}{2}}\\ &=\int \frac{d(tan\frac{x}{2})}{(tan\frac{x}{2}+\frac{1}{2})^2+\frac{3}{4}}\\ &=\int \frac{d(tan\frac{x}{2})}{\frac{3}{4}[(\frac{2}{\sqrt{3}}tan\frac{x}{2}+\frac{1}{\sqrt{3}})^2+1]}\\ &=\frac{2\sqrt{3}}{3}\int \frac{d(\frac{2}{\sqrt{3}}tan\frac{x}{2}+\frac{1}{\sqrt{3}})}{(\frac{2}{\sqrt{3}}tan\frac{x}{2}+\frac{1}{\sqrt{3}})^2+1}\\ &=\frac{2\sqrt{3}}{3}arctan(\frac{2}{\sqrt{3}}tan\frac{x}{2}+\frac{1}{\sqrt{3}})+C \end{aligned}$

$\begin{aligned} 17. \int \frac{dx}{1+sinx+cosx}&=\int \frac{dx}{1+2sin\frac{x}{2}cos\frac{x}{2}+2cos^2\frac{x}{2}-1}\\ &=\int \frac{d(\frac{x}{2})}{sin\frac{x}{2}cos\frac{x}{2}+cos^2\frac{x}{2}}\\ &=\int \frac{sec^2\frac{x}{2}d(\frac{x}{2})}{tan\frac{x}{2}+1}\\ &=\int \frac{d(tan\frac{x}{2}+1)}{tan\frac{x}{2}+1}\\ &=ln\begin{vmatrix}tan\frac{x}{2}+1\end{vmatrix}+C \end{aligned}$

$\begin{aligned} 18. \int \frac{dx}{2sinx-cosx+5}&\end{aligned}$

令 $t=tan\frac{x}{2}$ ，则 $dx=\frac{2dt}{1+t^2}$ ， $sinx=\frac{2t}{1+t^2}$ ， $cosx=\frac{1-t^2}{1+t^2}$

$\begin{aligned} \therefore \int \frac{dx}{2sinx-cosx+5}&=\int \frac{1}{\frac{4t}{1+t^2}-\frac{1-t^2}{1+t^2}+5}\cdot \frac{2dt}{1+t^2}\\ &=\frac{1}{3}\int \frac{1}{(1+\frac{1}{3})^2+(\frac{\sqrt{5}}{3})^2}d(t+\frac{1}{3})\\ &=\frac{1}{\sqrt{5}}arctan\frac{3t+1}{\sqrt{5}}+C\\ &=\frac{1}{\sqrt{5}}arctan\frac{3tan\frac{x}{2}+1}{\sqrt{5}}+C \end{aligned}$

$\begin{aligned} 19. \int \frac{dx}{1+\sqrt[3]{x+1}} \end{aligned}$

令 $t=\sqrt[3]{x+1}$ ，则 $x=t^3-1$

$\begin{aligned} \therefore \int \frac{dx}{1+\sqrt[3]{x+1}}&=\int \frac{3t^2dt}{1+t}\\ &=3\int \frac{t^2-1+1}{1+t}dt\\ &=3\int (t-1)dt+3\int \frac{1}{t+1}dt\\ &=\frac{3}{2}t^2-3t+3ln|t+1|+C\\ &=\frac{3}{2}(x+1)^{\frac{2}{3}}-3\sqrt[3]{x+1}+3ln|\sqrt[3]{x+1}+1|+C \end{aligned}$

$\begin{aligned} 20. \int \frac{(\sqrt{x})^3-1}{\sqrt{x}+1}dx\end{aligned}$

令 $t=\sqrt{x}$ ，则

$\begin{aligned} \int \frac{(\sqrt{x})^3-1}{\sqrt{x}+1}dx&=\int \frac{t^3-1}{t+1}d(t^2)\\ &=2\int \frac{t(t^3+1-2)}{t+1}dt\\ &=2\int \frac{t(t+1)(t^2-t+1)}{t+1}dt-4\int \frac{t}{t+1}dt\\ &=2\int (t^3-t^2+t)dt-4\int \frac{t+1-1}{t+1}dt\\ &=\frac{t^4}{2}-\frac{2t^3}{3}+t^2-4t+4ln|t+1|+C\\ &=\frac{x^2}{2}-\frac{2x^{\frac{3}{2}}}{3}+x-4\sqrt{x}+4ln(\sqrt{x}+1)+C \end{aligned}$

$\begin{aligned} 21. \int \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}dx \end{aligned}$

令 $t=\sqrt{x+1}$ ，则 $x=t^2-1$

$\begin{aligned} \int \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}dx&=\int \frac{t-1}{t+1}d(t^2-1)\\ &=\int \frac{2t(t-1)}{t+1}dt\\ &=\int \frac{2t(t+1-2)}{t+1}dt\\ &=\int 2tdt-4\int \frac{t+1-1}{t+1}dt\\ &=t^2-4t+4ln|t+1|+C\\ &=x+1-4\sqrt{x+1}+4ln|\sqrt{x+1}+1|+C \end{aligned}$

$\begin{aligned} 22. \int \frac{dx}{\sqrt{x}+\sqrt[4]{x}}dx\end{aligned}$

令 $t=\sqrt[4]{x}$ ，则 $x=t^4$

$\begin{aligned} \int \frac{dx}{\sqrt{x}+\sqrt[4]{x}}&=\int \frac{4t^3dt}{t^2+t}\\ &=4\int \frac{t^2dt}{t+1}\\ &=4\int \frac{t^2-1+1}{t+1}dt\\ &=4\int (t-1)dt+4\int \frac{dt}{t+1}\\ &=2t^2-4t+4ln(t+1)+C\\ &=2\sqrt{x}-4\sqrt[4]{x}+4ln(\sqrt[4]{x}+1)+C \end{aligned}$