高等数学：不定积分题选(2)

作者: 溺于恐 | 来源:发表于2018-12-13 08:26 被阅读26次

1. $\int {x^3+1\over (x^2+1)^2}dx$

解：

$设x=tant,-{\pi\over 2}\lt t\lt {\pi\over 2},则dx=sec^2tdt$

$原式=\int {tan^3t+1\over (tan^2+1)^2}sec^2tdt$

$=\int {tan^3t+1\over sec^2t}dt$

$=\int {sin^3t+cos^3t\over cost}dt$

$=\int {sin^3t\over cost}dt+\int cos^2tdt$

$=\int {cos^2t-1\over cost}d(cost)+\int {1+cos2t\over 2}dt$

$={1\over 2}cos^2t-lncost+{1\over 2}t+{1\over 4}sin2t+C$

$={1\over 2(1+x^2)}-ln{1\over \sqrt{1+x^2}}+{1\over 2}arctanx+{x\over 2(1+x^2)}+C$

$={1+x\over 2(1+x^2)}+{1\over 2}ln(1+x^2)+{1\over 2}arctanx+C$

2. $\int x^2arctanxdx$

解：

$原式={1\over 3}\int arctanxd(x^3)$

$={1\over 3}(x^3arctanx-\int x^3{1\over 1+x^2}dx)$

$={1\over 3}x^3arctanx-{1\over 3}\int {x^3+x-x\over 1+x^2}dx$

$={1\over 3}x^3arctanx-{1\over 3}\int xdx+{1\over 3}\int {xdx\over 1+x^2}$

$={1\over 3}x^3arctanx-{1\over 6}x^2+{1\over 6}ln(1+x^2)+C$

2. $\int xtan^2xdx$

解：

$原式=\int x(sec^2x-1)dx$

$=\int xsec^2xdx-\int xdx$

$=\int xd(tanx)-{1\over 2}x^2$

$=xtanx-\int tanxdx-{1\over 2}x^2$

$=xtanx+ln|cosx|-{1\over 2}x^2+C$

3. $\int x^2cosxdx$

解：

$原式=\int x^2d(sinx)$

$=x^2sinx-\int 2xsinxdx$

$=x^2sinx+2\int xd(cosx)$

$=x^2sinx+2xcosx-2\int cosxdx$

$=x^2sinx+2xcosx-2sinx+C$

4. $\int xsinxcosxdx$

解：

$原式={1\over 2}\int sin2xdx$

$=-{1\over 4}\int xd(cos2x)$

$=-{1\over 4}(xcos2x-\int cos2xdx)$

$=-{1\over 4}xcos2x+{1\over 8}sin2x+C$

5. $\int x^2cos^2{x\over 2}dx$

解：

$原式={1\over 2}\int x^2(1+cosx)dx$

$={1\over 2}\int x^2+{1\over 2}\int x^2cosxdx$

$={1\over 6}x^3+{1\over 2}\int x^2d(sinx)$

$={1\over 6}x^3+{1\over 2}(x^2sinx-\int 2xsinxdx)$

$={1\over 6}x^3+{1\over 2}x^2sinx+\int xd(cosx)$

$={1\over 6}x^3+{1\over 2}x^2sinx+xcosx-\int cosxdx$

$={1\over 6}x^3+{1\over 2}x^2sinx+xcosx-sinx+C$

6. $\int xln(x-1)dx$

解：

$原式={1\over 2}\int ln(x-1)d(x^2)$

$={1\over 2}[x^2ln(x-1)-\int {x^2\over x-1}dx]$

$={1\over 2}[x^2ln(x-1)-\int {x^2-1+1\over x-1}dx]$

$={1\over 2}x^2ln(x-1)-{1\over 2}[\int (x+1)dx+\int {1\over x-1}dx]$

$={1\over 2}x^2ln(x-1)-{1\over 4}(x+1)^2-{1\over 2}ln(x-1)+C$

$={1\over 2}(x^2-1)ln(x-1)-{1\over 4}x^2-{1\over 2}x+C$

7. $\int (x^2-1)sin2xdx$

解：

$原式=\int x^2sin2xdx-\int sin2xdx$

$=-{1\over 2}\int x^2d(cos2x)+{1\over 2}cos2x$

$=-{1\over 2}(x^2cos2x-\int 2xcos2xdx)+{1\over 2}cos2x$

$=-{1\over 2}x^2cos2x+{1\over 2}\int xd(sin2x)+{1\over 2}cos2x$

$=-{1\over 2}x^2cos2x+{1\over 2}(xsin2x-\int sin2xdx)+{1\over 2}cos2x$

$=-{1\over 2}x^2cos2x+{1\over 2}xsin2x+{1\over 4}cos2x+{1\over 2}cos2x+C$
$=-{1\over 2}(x^2-{3\over 2})cos2x+{x\over 2}sin2x+C$

8. $\int {ln^3x\over x^2}dx$

解：

$原式=\int ln^3xd(-{1\over x})$

$=-{1\over x}ln^3x+\int {3ln^2x\over x^2}dx$

$=-{1\over x}ln^3x-3\int ln^2xd({1\over x})$

$=-{1\over x}ln^3x-3[{1\over x}ln^2x-\int {2lnx\over x^2}dx]$

$=-{1\over x}ln^3x-{3\over x}ln^2x-6\int lnxd({1\over x})$

$=-{1\over x}ln^3x-{3\over x}ln^2x-6[{1\over x}lnx-\int {1\over x^2}dx]$

$=-{1\over x}ln^3x-{3\over x}ln^2x-{6\over x}lnx-6{1\over x}+C$

$=-{1\over x}(ln^3x+3ln^2x+6lnx+6)+C$

9. $\int e^{\sqrt[3]{x}}dx$

解：

$设\sqrt[3]{x}=t,则x=t^3$

$原式=\int e^t3t^2dt$

$=3\int t^2d(e^t)$

$=3(t^2e^t-\int e^t2tdt)$

$=3t^2e^t-6\int td(e^t)$

$=3t^2e^t-6te^t+6\int e^tdt$

$=3t^2e^t-6te^t+6e^t+C$

$=3e^{\sqrt[3]{x}}(\sqrt[3]{x^2}-2\sqrt[3]{x}+2)+C$

10. $\int (arcsinx)^2dx$

解：

$原式=x(arcsinx)^2-\int {x2arcsinx\over \sqrt{1-x^2}}dx$

$=x(arcsinx)^2+\int {arcsinx\over \sqrt{1-x^2}}d(1-x^2)$

$=x(arcsinx)^2+\int 2arcsinxd\sqrt{1-x^2}$

$=x(arcsinx)^2+2(\sqrt{1-x^2}arcsinx-\int \sqrt{1-x^2}{1\over \sqrt{1-x^2}}dx)$

$=x(arcsinx)^2+2\sqrt{1-x^2}arcsinx-2x+C$

11. $\int e^xsin^2xdx$

解：

$原式=\int e^x{1-cos2x\over 2}dx$

$={1\over 2}e^x-{1\over 2}\int e^xcos2xdx$

$设I=\int e^xcos2xdx$

$则I=\int cos2xd(e^x)$

$=e^xcos2x-\int e^xd(cos2x)$

$=e^xcos2x+2\int e^xsin2xdx$

$=e^xcos2x+2\int sin2xd(e^x)$

$=e^xcos2x+2sin2xe^x-2\int e^xd(sin2x)$

$=e^xcos2x+2sin2xe^x-4\int cos2xe^xdx$

$=e^xcos2x+2sin2xe^x-4I$

$\therefore 5I=e^xcos2x+2sin2xe^x$

$I={1\over 5}e^xcos2x+{2\over 5}e^xsin2x+C_1$

$\therefore 原积分={1\over 2}e^x-{1\over 10}e^xcos2x-{1\over 5}e^xsin2x+C$

12. $\int xln^2xdx$

解：

$原式=\int ln^2xd({x^2\over 2})$

$={x^2\over 2}ln^2x-\int {x^2\over 2}2lnx\cdot{1\over x}dx$

$={x^2\over 2}ln^2x-\int xlnxdx$

$={x^2\over 2}ln^2x-\int lnxd({x^2\over 2})$

$={x^2\over 2}ln^2x-{x^2\over 2}lnx+\int {x\over 2}dx$

$={x^2\over 2}ln^2x-{x^2\over 2}lnx+{x^2\over 4}+C$

$={x^2\over 4}(2ln^2x-2lnx+1)+C$

13. $\int e^{\sqrt{3x+9}}dx$

解：

$设\sqrt{3x+9}=u,则x={1\over 3}(u^2-9)$

$原式=\int e^u{2\over 3}udu$

$=\int {2\over 3}ud(e^u)$

$={2\over 3}ue^u-{2\over 3}\int e^udu$

$={2\over 3}ue^u-{2\over 3}e^u+C$

$={2\over 3}e^{\sqrt{3x+9}}(\sqrt{3x+9}-1)+C$

14. $\int {x^3\over x+3}dx$

解：

$原式=\int {x^3+27-27\over x+3}dx$

$=\int {(x+3)(x^2-3x+9)-27\over x+3}dx$

$=\int (x^2-3x+9)dx-\int {27\over x+3}dx$

$={1\over 3}x^3-{3\over 2}x^2+9x-27ln|x+3|+C$

15. $\int {2x+3\over x^2+3x-10}dx$

解：

$原式=\int {d(x^2+3x-10)\over x^2+3x-10}$

$=ln|x^2+3x-10|+C$

$=ln|x-2|+ln|x+5|+C$

16. $\int {x+1\over x^2-2x+5}dx$

解：

$原式=\int {x-1+2\over (x-1)^2+4}dx$

$=\int {x-1\over (x-1)^2+4}dx+{1\over 2}\int {1\over ({x-1\over 2})^2+1}dx$

$={1\over 2}ln(x^2-2x+5)+{1\over 2}arctan{x-1\over 2}+C$

17. $\int {3\over x^3+1}dx$

解：

$原式=\int {3\over (x+1)(x^2-x+1)}dx$

$=\int ({1\over x+1}+{-x+2\over x^2-x+1})dx$

$=ln|x+1|+\int {-x+{1\over 2}\over x^2-x+1}dx+\int {{3\over 2}dx\over x^2-x+1}$

$=ln|x+1|-{1\over 2}\int {d(x^2-x+1)\over x^2-x+1}+{3\over 2}\int {d(x-{1\over 2})\over (x-{1\over 2})^2+({\sqrt{3}\over 2})^2}$

$=ln|x+1|-{1\over 2}ln|x^2-x+1|+{3\over 2}{2\over \sqrt{3}}arctan{x-{1\over 2}\over {\sqrt{3}\over 2}}$

$=ln{|x+1|\over \sqrt{x^2-x+1}}+\sqrt{3}arctan{2x-1\over \sqrt{3}}$

18. $\int {x^2+1\over (x+1)^2(x-1)}dx$

分析：

$因式分解$

$设{x^2+1\over (x+1)^2(x-1)}={A\over (x+1)^2}+{B\over x+1}+{C\over x-1}$

$即A(x-1)+B(x^2-1)+C(x+1)^2=x^2+1$

$整理得(B+C)x^2+(A+2C)x+C-A-B=x^2+1$

$比较系数得\begin{cases}B+C=1 \\A+2C=0 \\C-A-B=1\end{cases}$

$解得A=-1,B={1\over 2},C={1\over 2}$

$即{x^2+1\over (x+1)^2(x-1)}={-1\over (x+1)^2}+{1\over 2(x+1)}+{1\over 2(x-1)}$

解：

$原式=\int [{-1\over (x+1)^2}+{1\over 2(x+1)}+{1\over 2(x-1)}]dx$

$={1\over x+1}+{1\over 2}ln|x+1|+{1\over 2}ln|x-1|+C$

$={1\over x+1}+{1\over 2}ln|x^2-1|+C$

19. $\int {xdx\over (x+1)(x+2)(x+3)}$

分析：

$因式分解$

$设{x\over (x+1)(x+2)(x+3)}={A\over x+1}+{B\over x+2}+{C\over x+3}$

$即A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)=x$

$整理得(A+B+C)x^2+(5A+4B+3C)x+6A+3B+2C=x$

$比较系数得\begin{cases}A+B+C=0 \\5A+4B+3C=1 \\6A+3B+2C=0\end{cases}$

$解得A=-{1\over 2},B=2,C=-{3\over 2}$

$即{x\over (x+1)(x+2)(x+3)}=-{1\over 2(x+1)}+{2\over x+2}-{3\over 2(x+3)}$

解：

$原式=\int [-{1\over 2(x+1)}+{2\over x+2}-{3\over 2(x+3)}]dx$

$=-{1\over 2}ln|x+1|+2ln|x+2|-{3\over 2}ln|x+3|+C$

$={1\over 2}ln{(x+2)^4\over |x+1||x+3|^3}+C$

20. $\int {x^5+x^4-8\over x^3-x}dx$

分析：

$x^5+x^4-8=(x^2+x+1)(x^3-x)+x^2+x-8$

$即{x^5+x^4-8\over x^3-x}=x^2+x+1+{x^2+x-8\over x^3-x}$

$设{x^2+x-8\over x^3-x}={A\over x}+{B\over x-1}+{C\over x+1}$

$即A(x^2-1)+B(x^2+x)+C(x^2-x)=x^2+x-8$

$比较系数得\begin{cases}A+B+C=1 \\-A=-8 \\B-C=1\end{cases}$

$解得A=8,B=-3,C=-4$

$即{x^5+x^4-8\over x^3-x}=x^2+x+1+{8\over x}-{3\over x-1}-{4\over x+1}$

解：

$原式=\int (x^2+x+1+{8\over x}-{3\over x-1}-{4\over x+1})dx$

$={1\over 3}x^3+{1\over 2}x^2+x+8ln|x|-3ln|x-1|-4ln|x+1|+C$

21. $\int {dx\over (x^2+1)(x^2+x)}$

分析：

$设{1\over (x^2+1)(x^2+x)}={A\over x}+{B\over x+1}+{Cx+D\over x^2+1}$

$即A(x+1)(x^2+1)+Bx(x^2+1)+x(x+1)(Cx+D)=1$

$比较系数得\begin{cases}A+B+C=0 \\A+C+D=0 \\A+B+D=0 \\A=1\end{cases}$

$解得A=1,B=-{1\over 2},C=-{1\over 2},D=-{1\over 2}$

解：

$原式=\int ({1\over x}-{1\over 2(x+1)}-{1+x\over 2(x^2+1)})dx$

$=ln|x|-{1\over 2}ln|x+1|-{1\over 4}ln|x^2+1|-{1\over 2}arctanx+C$

22. $\int {1\over x^4-1}dx$

分析：

$设{1\over (x-1)(x+1)(x^2+1)}={A\over x-1}+{B\over x+1}+{C\over x^2+1}$

$即A(x+1)(x^2+1)+B(x-1)(x^2+1)+C(x^2-1)=1$

$比较系数得\begin{cases}A+B=0 \\A-B+C=0 \\A-B-C=1\end{cases}$

$解得A={1\over 4},B=-{1\over 4},C=-{1\over 2}$

解：

$原式=\int {dx\over (x-1)(x+1)(x^2+1)}$

$={1\over 4}\int {dx\over x-1}-{1\over 4}\int {dx\over x+1}-{1\over 2}\int {dx\over x^2+1}$

$={1\over 4}ln|x-1|-{1\over 4}ln|x+1|-{1\over 2}arctanx+C$

$={1\over 4}ln|{x-1\over x+1}|-{1\over 2}arctanx+ C$

23. $\int {dx\over (x^2+1)(x^2+x+1)}$

分析：

$设{1\over (x^2+1)(x^2+x+1)}={Ax\over x^2+1}+{Bx+C\over x^2+x+1}$

$即Ax(x^2+x+1)+(Bx+C)(x^2+1)=1$

$比较系数得\begin{cases}A+B=0 \\A+C=0 \\C=1\end{cases}$

$解得A=-1,B=1,C=1$

解：

$原式=\int ({-x\over x^2+1}+{x+1\over x^2+x+1})dx$

$=-{1\over 2}\int ({dx^2\over x^2+1}+{x+{1\over 2}\over x^2+x+1}+{{1\over 2}\over x^2+x+1})dx$

$=-{1\over 2}\int {d(x^2+1)\over x^2+1}+{1\over 2}{d(x^2+x+1)\over x^2+x+1}+{1\over 2}\int {dx\over (x+{1\over 2})^2+({\sqrt{3}\over 2})^2}$

$=-{1\over 2}ln(x^2+1)+{1\over 2}ln(x^2+x+1)+{1\over 2}\times {2\over \sqrt{3}}arctan{x+{1\over 2}\over {\sqrt{3}\over 2}}+C$

$={1\over 2}ln{x^2+x+1\over x^2+1}+{\sqrt{3}\over 3}arctan{2x+1\over \sqrt{3}}+C$

24. $\int {(x+1)^2\over (x^2+1)^2}dx$

解：

$原式=\int {x^2+1+2x\over (x^2+1)^2}dx$

$=\int {1\over x^2+1}dx+\int{2x\over (x^2+1)^2}dx$

$=arctanx+\int {d(x^2+1)\over (x^2+1)^2}$

$=arctanx-{1\over x^2+1}+C$

25. $\int {-x^2-2\over (x^2+x+1)^2}dx$

解：

$原式=\int {-x^2-x-1+x-1\over (x^2+x+1)^2}dx$

$=-\int {dx\over x^2+x+1}+\int {x-1\over (x^2+x+1)^2}dx$

$=-\int {dx\over (x+{1\over 2})^2+{3\over 4}}+\int {x+{1\over 2}\over (x^2+x+1)^2}dx-\int {{3\over 2}\over (x^2+x+1)^2}dx$

$=-{2\over \sqrt{3}}arctan{x+{1\over 2}\over {\sqrt{3}\over 2}}+{1\over 2}\int {d(x^2+x+1)\over (x^2+x+1)^2}-{3\over 2}\int {d(x+{1\over 2})\over [(x+{1\over 2})^2+{3\over 4}]^2}$

$=-{2\over \sqrt{3}}arctan{2x+1\over \sqrt{3}}-{1\over 2}{1\over x^2+x+1}-{3\over 2}\times {1\over 2({\sqrt{3}\over 2})^2}\times [{x+{1\over 2}\over (x+{1\over 2})^2+({\sqrt{3}\over 2})^2}+{2\over \sqrt{3}}arctan{2x+1\over \sqrt{3}}]+C$