7.3 讲义习题

7.3.1 不定积分的运算性质

解 (1) $$\begin{array}{}\text{(1)}& \int x\sqrt{x\sqrt{x}}\mathrm{d}x=\int x^{7/4}\mathrm{d}x=\frac{4}{11}{x}^{11/4}+C\end{array}$$

(2) $$\begin{array}{}\text{(2)}& \int \frac{1}{x^2}\cos \frac{1}{x}\mathrm{d}x=-\int \cos \frac{1}{x}\mathrm{d}\left(\frac{1}{x}\right)=-\sin \frac{1}{x}+C\end{array}$$

(3) $$\begin{array}{}\text{(3)}& \int \frac{1}{{\mathrm{e}}^x+{\mathrm{e}}^{-x}}\mathrm{d}x=\int \frac{\mathrm{d}({\mathrm{e}}^x)}{{\mathrm{e}}^{2x}+1}=\arctan ({\mathrm{e}}^x)+C\end{array}$$

(4) 设$x\in [-\frac{\pi }{2},\frac{\pi }{2}]$，则$$\begin{array}{}\text{(4)}& \int \sqrt{1+\cos 2x}\mathrm{d}x=\int \sqrt{2}\cos x\mathrm{d}x=\sqrt{2}\sin x+C\end{array}$$

(5) $$\begin{array}{}\text{(5)}& \int \frac{\mathrm{d}x}{\sqrt{1+x^2}}\stackrel{x=\tan t}{=}\int \sec t\mathrm{d}t=\ln |\sec t+\tan t|+C=\ln (x+\sqrt{1+x^2})+C\end{array}$$ $◻$

解 (1) 设$[{\mathrm{e}}^{\lambda x}(A\sin x+B\cos x){]}^{\prime }={\mathrm{e}}^{\lambda x}\sin x$，则$$\begin{array}{}\text{(6)}& {\mathrm{e}}^{\lambda x}\sin x={\mathrm{e}}^{\lambda x}[(\lambda A-B)\sin x+(A+\lambda B)\cos x]⟹A=\frac{\lambda }{{\lambda }^2+1},B=-\frac{1}{{\lambda }^2+1}\end{array}$$ 因此$$\begin{array}{}\text{(7)}& \int {\mathrm{e}}^{\lambda x}\sin x\mathrm{d}x=\frac{{\mathrm{e}}^{\lambda x}(\lambda \sin x-\cos x)}{{\lambda }^2+1}+C\end{array}$$

(2) 设${\alpha }^2\ne {\beta }^2$，则$$\begin{array}{}\text{(8)}& \begin{array}{rl}\int \sin \alpha x\cos \beta x\mathrm{d}x& =\frac{1}{2}\int [\sin (\alpha +\beta )x+\sin (\alpha -\beta )x]\mathrm{d}x\\ & =-\frac{\cos (\alpha +\beta )x}{2(\alpha +\beta )}-\frac{\cos (\alpha -\beta )x}{2(\alpha -\beta )}+C\end{array}\end{array}$$

(3) $$\begin{array}{}\text{(9)}& \int \tan x\mathrm{d}x=-\int \frac{\mathrm{d}(\cos x)}{\cos x}=-\ln |\cos x|+C\end{array}$$

(4) 设第一、二个被积函数的原函数分别为$F_1,F_2$，则$$\begin{array}{}\text{(10)}& \begin{array}{rl}a{F}_1+b{F}_2& =\int \mathrm{d}x=x+C_1,\\ -b{F}_1+a{F}_2& =\int \frac{a\cos x-b\sin x}{a\sin x+b\cos x}\mathrm{d}x=\ln |a\sin x+b\cos x|+C_2\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(11)}& \begin{array}{rl}\int \frac{\sin x\mathrm{d}x}{a\sin x+b\cos x}& =\frac{ax-b\ln |a\sin x+b\cos x|}{a^2+b^2}+C\\ \int \frac{\cos x\mathrm{d}x}{a\sin x+b\cos x}& =\frac{bx+a\ln |a\sin x+b\cos x|}{a^2+b^2}+C\end{array}\end{array}$$

(5) $$\begin{array}{}\text{(12)}& \int x^4{\ln }^{2}x\mathrm{d}x=\frac{1}{5}{x}^5{\ln }^{2}x-\frac{2}{5}\int x^4\ln x\mathrm{d}x=\frac{1}{5}{x}^5{\ln }^{2}x-\frac{2}{25}{x}^5\ln x+\frac{2}{125}{x}^5+C\end{array}$$

(6) $$\begin{array}{}\text{(13)}& \int \sin (\ln x)\mathrm{d}x\stackrel{x={\mathrm{e}}^t}{=}\int {\mathrm{e}}^t\sin t\mathrm{d}t=\frac{1}{2}x(\sin (\ln x)-\cos (\ln x))+C\end{array}$$ $◻$

7.3.2 有理函数的不定积分以及可转化为有理函数的不定积分

解 (1) 设$$\begin{array}{}\text{(14)}& \frac{1}{x^2(1+x^2{)}^2}=\frac{A}{x^2}+\frac{E}{1+x^2}+\frac{F}{(1+x^2{)}^2}⟹1=A(1+x^2{)}^2+E{x}^2(1+x^2)+F{x}^2\end{array}$$ 令$x=0$可得$A=1$，令$x=\mathrm{i}$可得$F=-1$。等式两边同时求导并代入$x=\mathrm{i}$可得$$\begin{array}{}\text{(15)}& 0=0+0+2E{x}^3+2Fx⟹E=F=-1\end{array}$$ 因此$$\begin{array}{}\text{(16)}& \int \frac{\mathrm{d}x}{x^2(1+x^2{)}^2}=-\frac{1}{x}-\frac{x}{2(1+x^2)}-\frac{3}{2}\arctan x+C\end{array}$$ 其中$$\begin{array}{}\text{(17)}& \begin{array}{rl}{I}_n:=& \int \frac{\mathrm{d}x}{(1+x^2{)}^n}=\frac{x}{(1+x^2{)}^n}+2n\int \frac{x^2\mathrm{d}x}{(1+x^2{)}^{n+1}}=\frac{x}{(1+x^2{)}^n}+2n(I_n-I_{n+1})\\ & ⟹I_{n+1}=\frac{x}{2n(1+x^2{)}^n}+\frac{2n-1}{2n}{I}_n\end{array}\end{array}$$

(2) 设$$\begin{array}{}\text{(18)}& \frac{x^3}{(1-x^2{)}^3}=\frac{Ax}{1-x^2}+\frac{Bx}{(1-x^2{)}^2}+\frac{Dx}{(1-x^2{)}^3}⟹x^2=A(1-x^2{)}^2+B(1-x^2)+D\end{array}$$ 令$x=1$可得$D=1$。等式两边同时求导可得$$\begin{array}{}\text{(19)}& 2x=-4Ax(1-x^2)-2Bx⟹A=0,B=-1\end{array}$$ 因此$$\begin{array}{}\text{(20)}& \int \frac{x^3}{(1-x^2{)}^3}\mathrm{d}x=-\frac{1}{2(1-x^2)}+\frac{1}{4(1-x^2{)}^2}+x+C\end{array}$$

(3) 令$t=\sqrt[3]{\frac{x+3}{1-x}}$，则$x=\frac{t^3-3}{t^3+1}$，$\mathrm{d}x=\frac{12t^2}{(1+t^3{)}^2}\mathrm{d}t$，此时$$\begin{array}{}\text{(21)}& \begin{array}{rl}& \int \sqrt[3]{\frac{x+3}{1-x}}\mathrm{d}x=12\int \frac{t^3\mathrm{d}t}{(1+t^3{)}^2}=-\frac{4t}{1+t^3}+\frac{4\sqrt{3}}{3}\arctan \frac{2t-1}{\sqrt{3}}+\frac{2}{3}\ln \frac{(t+1{)}^2}{t^2-t+1}+C\\ & =(x-1)\sqrt[3]{\frac{x+3}{1-x}}+\frac{4\sqrt{3}}{3}\arctan \frac{2\sqrt[3]{\frac{x+3}{1-x}}-1}{\sqrt{3}}+2\ln (1+\sqrt[3]{\frac{x+3}{1-x}})+\frac{2}{3}\ln (1-x)+C\end{array}\end{array}$$

(4) $$\begin{array}{}\text{(22)}& \int \sqrt{1+x^2}\mathrm{d}x\stackrel{x=\tan t}{=}\int {\sec }^{3}t\mathrm{d}t=\frac{1}{2}[x\sqrt{1+x^2}+\ln (x+\sqrt{1+x^2})]+C\end{array}$$

(5) $$\begin{array}{}\text{(23)}& \int \frac{1}{\sqrt{x}+\sqrt[3]{x}}\mathrm{d}x\stackrel{x=t^6}{=}\int \frac{6t^5}{t^3+t^2}\mathrm{d}t=6\sqrt[6]{x}-3\sqrt[3]{x}+2\sqrt{x}-6\ln (1+\sqrt[6]{x})+C\end{array}$$

(6) 令$t=x+1$，则$$\begin{array}{}\text{(24)}& \begin{array}{rl}& \int \frac{x^2-x+1}{\sqrt{x^2+2x+2}}\mathrm{d}x=\int \frac{t^2+1-3t+2}{\sqrt{t^2+1}}\mathrm{d}t=\int \sqrt{1+t^2}\mathrm{d}t+\int \frac{2\mathrm{d}t}{\sqrt{1+t^2}}-3\sqrt{1+t^2}\\ & =\frac{1}{2}[t\sqrt{1+t^2}+\ln (t+\sqrt{1+t^2})]+2\ln (t+\sqrt{1+t^2})-3\sqrt{1+t^2}+C\\ & =\frac{1}{2}(x+1)\sqrt{x^2+2x+2}+\frac{5}{2}\ln (x+1+\sqrt{x^2+2x+2})-3\sqrt{x^2+2x+2}+C\end{array}\end{array}$$

(7) 令$x=\sin t$、$u=\cos t=\sqrt{1-x^2}$，则$$\begin{array}{}\text{(25)}& \begin{array}{rl}\int x^3\sqrt{1-x^2}\mathrm{d}x& =\int {\sin }^{3}t{\cos }^{2}t\mathrm{d}t=\int (u^2-1)u^2\mathrm{d}u=\frac{1}{5}{(1-x^2)}^{5/2}-\frac{1}{3}{(1-x^2)}^{3/2}+C\end{array}\end{array}$$

(8) $$\begin{array}{}\text{(26)}& \begin{array}{rl}\int {\sec }^{3}x\mathrm{d}x& =\int \sec x\mathrm{d}(\tan x)=\sec x\tan x-\int \sec x{\tan }^{2}x\mathrm{d}x\\ & =\sec x\tan x+\int \sec x\mathrm{d}x-\int {\sec }^{3}x\mathrm{d}x\\ & =\frac{1}{2}(\sec x\tan x+\ln |\sec x+\tan x|)+C\end{array}\end{array}$$

(9) 令$t=\cos x$，则$$\begin{array}{}\text{(27)}& \int \frac{{\sin }^{5}x}{{\cos }^{4}x}\mathrm{d}x=-\int \frac{(1-t^2{)}^2\mathrm{d}t}{t^4}=\frac{1}{3{\cos }^{3}x}-\frac{2}{\cos x}-\cos x+C\end{array}$$

(10) 令$t=\sqrt[3]{1+x^5}$，则$\frac{\mathrm{d}x}{x}=\frac{5x^4\mathrm{d}x}{5x^5}=\frac{3t^2\mathrm{d}t}{5(t^3-1)}$$$\begin{array}{}\text{(28)}& \int \frac{\mathrm{d}x}{x\sqrt[3]{1+x^5}}=\frac{3}{5}\int \frac{t\mathrm{d}t}{t^3-1}=\frac{\sqrt{3}}{5}\arctan \frac{1+2\sqrt[3]{1+x^5}}{\sqrt{3}}+\frac{3}{10}\ln |\sqrt[3]{1+x^5}-1|-\frac{1}{2}\ln |x|+C\end{array}$$

(11) 利用万能公式可得$$\begin{array}{}\text{(29)}& \int \frac{\mathrm{d}x}{3+2\sin x}=\frac{2}{\sqrt{5}}\arctan \frac{2+3\tan \frac{x}{2}}{\sqrt{5}}+C\end{array}$$

(12) $$\begin{array}{}\text{(30)}& \int {\sec }^{4}x\mathrm{d}x=\int (1+{\tan }^{2}x)\mathrm{d}(\tan x)=\tan x+\frac{1}{3}{\tan }^{3}x+C\end{array}$$

(13) 令$t=x^{-4}$、$u=\sqrt[4]{1+t}$，则$$\begin{array}{}\text{(31)}& \int \frac{\mathrm{d}x}{\sqrt[4]{1+x^4}}=-\frac{1}{4}\int \frac{\mathrm{d}t}{t\sqrt[4]{1+t}}=-\int \frac{u^2\mathrm{d}u}{u^4-1}=\frac{1}{2}\arctan \frac{x}{\sqrt[4]{1+x^4}}+\frac{1}{4}\ln \frac{x+\sqrt[4]{1+x^4}}{x-\sqrt[4]{1+x^4}}+C\end{array}$$

(14) 令$t=\sqrt[3]{1+\sqrt[4]{x}}$，则$$\begin{array}{}\text{(32)}& \int \frac{\sqrt[3]{1+\sqrt[4]{x}}}{\sqrt{x}}\mathrm{d}x=12\int (t^3-1)t^3\mathrm{d}t=\frac{12}{7}{(1+\sqrt[4]{x})}^{7/3}-3{(1+\sqrt[4]{x})}^{4/3}+C\end{array}$$ $◻$