高数之不定积分

文章目录

  • 一、第一类换元法
    • 1.核心公式
    • 2.基本积分
      • (1)分式
      • (2)三角函数
    • 3.写法
      • (1)凑微分法
      • (2)常见凑微分形式:
    • 4.题
      • (1)第一类换元法的基本应用
      • (2)第一类换元法和分式加减分解
      • (3)三角函数专题
  • 二、第二类换元法
    • 1.核心公式
    • 2.常用三种变量代换
      • (1)知识点
      • (2)题
    • 3.出错点
    • 4.题
      • (1)第二类换元法的基本应用
      • (2)难题
  • 三、分部积分法
    • 1.公式
    • 2.解析


一、第一类换元法

1.核心公式

∫ f [ ϕ ( x ) ] ϕ ′ ( x ) d x = ∫ f ( u ) d u ∣ u = ϕ ( x ) = F ( u ) + C \int f[\phi(x)]\phi'(x)dx=\int f(u)du|_{u=\phi(x)}=F(u)+C f[ϕ(x)]ϕ(x)dx=f(u)duu=ϕ(x)=F(u)+C

【思路】:

合并,从整体的角度将其化成基本积分解决。

【用法】:

找到可以化为基本积分形式的 f ( u ) f(u) f(u),然后变换积分部分构造成满足 d u du du的样子。

比如: ∫ 4 x e x 2 d x \displaystyle \int 4xe^{x^2}dx 4xex2dx e x 2 e^{x^2} ex2就是 f ( u ) f(u) f(u),那么其他部分变换成 d u du du的样子, d u = d ( x 2 ) = 2 x d x du=d(x^2)=2xdx du=d(x2)=2xdx,所以 ∫ 4 x e x 2 d x = ∫ 2 e x 2 d ( x 2 ) = 2 e x 2 + C \displaystyle \int 4xe^{x^2}dx=\displaystyle \int 2e^{x^2}d(x^2)=2e^{x^2}+C 4xex2dx=2ex2d(x2)=2ex2+C

2.基本积分

完整版:
https://wenku.baidu.com/view/8eeeeb6c561252d380eb6e08.html
https://blog.csdn.net/yellow_hill/article/details/82977844

(1)分式

【最基本的ln】

  • ∫ 1 x d x = l n ∣ x ∣ + C \displaystyle \int \dfrac{1}{x}dx= ln|x|+C x1dx=lnx+C
    记着绝对值
    x x x要是负的还得对 − x -x x求导得到 − 1 -1 1,如 ∫ 1 a − x d x = − ln ⁡ ( a − x ) d x \displaystyle \int \dfrac{1}{a-x}dx=-\ln{(a-x)} dx ax1dx=ln(ax)dx

【反三角函数】

  • ∫ 1 1 − x 2 d x = arcsin ⁡ x + C \displaystyle \int \dfrac{1}{\sqrt{1-x^2}}dx=\arcsin{x}+C 1x2 1dx=arcsinx+C
    ∫ 1 a 2 − x 2 d x = arcsin ⁡ x a + C ( a = ̸ 0 ) \displaystyle \int \dfrac{1}{\sqrt{a^2-x^2}}dx=\arcsin{\frac{x}{a}}+C \quad (a=\not 0) a2x2 1dx=arcsinax+C(a≠0)

  • ∫ 1 1 + x 2 d x = arctan ⁡ x + C \displaystyle \int \dfrac{1}{1+x^2}dx=\arctan{x}+C 1+x21dx=arctanx+C
    ∫ 1 a 2 + x 2 d x = 1 a arctan ⁡ x a + C ( a = ̸ 0 ) \displaystyle \int \dfrac{1}{a^2+x^2}dx=\frac{1}{a}\arctan{\frac{x}{a}}+C \quad (a=\not 0) a2+x21dx=a1arctanax+C(a≠0)

x 2 ± a 2 x^2 \pm a^2 x2±a2

  • ∫ 1 x 2 − a 2 d x = ∫ 1 x − a 1 x + a d x = ∫ 1 2 a ( 1 x − a − 1 x + a ) d x = 1 2 a ( ln ⁡ ∣ x − a ∣ − ln ⁡ ∣ x + a ∣ ) + C = 1 2 a ln ⁡ ∣ x − a x + a ∣ + C \displaystyle \int \dfrac{1}{x^2-a^2}dx=\displaystyle \int \dfrac{1}{x-a} \dfrac{1}{x+a} dx=\displaystyle \int \dfrac{1}{2a} (\dfrac{1}{x-a} -\dfrac{1}{x+a})dx=\dfrac{1}{2a}(\ln|x-a|-\ln|x+a|)+C=\dfrac{1}{2a}\ln|{\dfrac{x-a}{x+a}}|+C x2a21dx=xa1x+a1dx=2a1(xa1x+a1)dx=2a1(lnxalnx+a)+C=2a1lnx+axa+C

  • ∫ 1 a 2 − x 2 d x = − ∫ 1 x 2 − a 2 d x = ∫ 1 2 a ( − 1 x − a + 1 x + a ) d x = 1 2 a ( − ln ⁡ ∣ x − a ∣ + ln ⁡ ∣ x + a ∣ ) + C = 1 2 a ln ⁡ ∣ x + a x − a ∣ + C \displaystyle \int \dfrac{1}{a^2-x^2}dx =-\displaystyle \int \dfrac{1}{x^2-a^2}dx =\displaystyle \int \dfrac{1}{2a} (-\dfrac{1}{x-a} +\dfrac{1}{x+a})dx =\dfrac{1}{2a}(-\ln|x-a|+\ln|x+a|)+C =\dfrac{1}{2a}\ln|{\dfrac{x+a}{x-a}}|+C a2x21dx=x2a21dx=2a1(xa1+x+a1)dx=2a1(lnxa+lnx+a)+C=2a1lnxax+a+C

有请智商下线的沙雕推演:

  • ∫ 1 a 2 − x 2 d x = ∫ 1 a − x 1 a + x d x = ∫ 1 2 a ( 1 a − x + 1 a + x ) d x = ̸ 1 2 a ( ln ⁡ ∣ a − x ∣ + ln ⁡ ∣ a + x ∣ ) + C = 1 2 a ln ⁡ ( ∣ a − x ∣ ∣ a + x ∣ ) + C \displaystyle \int \dfrac{1}{a^2-x^2}dx =\displaystyle \int \dfrac{1}{a-x} \dfrac{1}{a+x} dx =\displaystyle \int \dfrac{1}{2a} (\dfrac{1}{a-x} +\dfrac{1}{a+x})dx =\not \dfrac{1}{2a}(\ln|a-x|+\ln|a+x|)+C =\dfrac{1}{2a}\ln(|a-x||a+x|)+C a2x21dx=ax1a+x1dx=2a1(ax1+a+x1)dx≠2a1(lnax+lna+x)+C=2a1ln(axa+x)+C
    注意:
    ∫ 1 a − x d x = − ln ⁡ ( a − x ) d x \displaystyle \int \dfrac{1}{a-x}dx=-\ln{(a-x)} dx ax1dx=ln(ax)dx,不是 = ln ⁡ ( a − x ) d x =\ln{(a-x)}dx =ln(ax)dx
    所以正确是
    ∫ 1 a 2 − x 2 d x = 1 2 a ( − ln ⁡ ∣ a − x ∣ + ln ⁡ ∣ a + x ∣ ) + C = 1 2 a ln ⁡ ∣ x + a x − a ∣ + C \displaystyle \int \dfrac{1}{a^2-x^2}dx =\dfrac{1}{2a}(-\ln|a-x|+\ln|a+x|)+C =\dfrac{1}{2a}\ln|{\dfrac{x+a}{x-a}}|+C a2x21dx=2a1(lnax+lna+x)+C=2a1lnxax+a+C

x 2 ± a 2 \sqrt{x^2 \pm a^2} x2±a2

  • ∫ 1 x 2 + a 2 d x = ln ⁡ ( x + x 2 + a 2 ) + C \displaystyle \int \dfrac{1}{\sqrt{x^2+a^2}}dx=\ln(x+\sqrt{x^2+a^2})+C x2+a2 1dx=ln(x+x2+a2 )+C

  • ∫ 1 x 2 − a 2 d x = ln ⁡ ( x + x 2 − a 2 ) + C \displaystyle \int \dfrac{1}{\sqrt{x^2-a^2}}dx=\ln(x+\sqrt{x^2-a^2})+C x2a2 1dx=ln(x+x2a2 )+C

(2)三角函数

  • ∫ 1 sin ⁡ 2 x d x = ∫ csc ⁡ 2 x d x = − cot ⁡ x + C \displaystyle \int \dfrac{1}{\sin^2{x}}dx=\displaystyle \int \csc^2{x}dx=-\cot{x}+C sin2x1dx=csc2xdx=cotx+C

  • ∫ 1 cos ⁡ 2 x d x = ∫ sec ⁡ 2 x d x = tan ⁡ x + C \displaystyle \int \dfrac{1}{\cos^2{x}}dx=\displaystyle \int \sec^2{x}dx=\tan{x}+C cos2x1dx=sec2xdx=tanx+C

  • ∫ sec ⁡ x d x = ln ⁡ ∣ sec ⁡ x + tan ⁡ x ∣ + C \displaystyle \int \sec{x}dx=\ln|\sec{x}+\tan{x}|+C secxdx=lnsecx+tanx+C

  • ∫ sec ⁡ 2 x d x = tan ⁡ x + C \displaystyle \int \sec^2{x}dx=\tan{x}+C sec2xdx=tanx+C

  • ∫ csc ⁡ x = ln ⁡ ∣ csc ⁡ x − cot ⁡ x ∣ + C = ln ⁡ ∣ tan ⁡ x x ∣ + C \displaystyle \int \csc{x}=\ln|\csc{x}-\cot{x}|+C=\ln|\tan{\frac{x}{x}}|+C cscx=lncscxcotx+C=lntanxx+C

  • ∫ csc ⁡ 2 x = − cot ⁡ x + C \displaystyle \int \csc^2{x}=-\cot{x}+C csc2x=cotx+C

  • ∫ tan ⁡ x = − ln ⁡ ∣ cos ⁡ x ∣ + C \displaystyle \int \tan{x}=-\ln|\cos{x}|+C tanx=lncosx+C

  • ∫ cot ⁡ x = ln ⁡ ∣ sin ⁡ x ∣ + C \displaystyle \int \cot{x}=\ln|\sin{x}|+C cotx=lnsinx+C

  • ∫ sec ⁡ x tan ⁡ x d x = sec ⁡ x + C \displaystyle \int \sec{x} \tan{x}dx=\sec{x}+C secxtanxdx=secx+C

  • ∫ csc ⁡ x cot ⁡ x d x = − csc ⁡ x + C \displaystyle \int \csc{x} \cot{x}dx=-\csc{x}+C cscxcotxdx=cscx+C

3.写法

(1)凑微分法

用第一类换元法,那么我们就要想法设法凑成 d ( ϕ ( x ) ) d(\phi(x)) d(ϕ(x))以便代入基本积分,所以叫凑微分法。
如: ∫ 2 x e x 2 d x = ∫ e x 2 d ( x 2 ) = e x 2 + C \displaystyle \int 2xe^{x^2}dx=\displaystyle \int e^{x^2}d(x^2)=e^{x^2}+C 2xex2dx=ex2d(x2)=ex2+C

(2)常见凑微分形式:

  1. ∫ f ( a x + b ) d x = 1 a ∫ f ( a x + b ) d ( a x + b ) \displaystyle \int f(ax+b)dx =\dfrac{1}{a} \displaystyle \int f(ax+b)d(ax+b) f(ax+b)dx=a1f(ax+b)d(ax+b)

  2. ∫ x m f ( a x m + 1 + b ) d x = 1 a ( m + 1 ) ∫ f ( a x m + 1 + b ) d ( a x m + 1 + b ) ( m = ̸ − 1 ) \displaystyle \int x^mf(ax^{m+1}+b)dx =\dfrac{1}{a(m+1)} \displaystyle \int f(ax^{m+1}+b)d(ax^{m+1}+b) \quad (m=\not -1) xmf(axm+1+b)dx=a(m+1)1f(axm+1+b)d(axm+1+b)(m≠1)

  3. ∫ f ( x ) 1 x d x = 2 ∫ f ( x ) d ( x ) \displaystyle \int f(\sqrt{x})\dfrac{1}{\sqrt{x}}dx =2\displaystyle \int f(\sqrt{x})d(\sqrt{x}) f(x )x 1dx=2f(x )d(x )

  4. ∫ f ( e x ) e x d x = ∫ f ( e x ) d ( e x ) \displaystyle \int f(e^x)e^xdx =\displaystyle \int f(e^x)d(e^x) f(ex)exdx=f(ex)d(ex)

  5. ∫ f ( ln ⁡ x ) 1 x d x = ∫ f ( ln ⁡ x ) d ( ln ⁡ x ) \displaystyle \int f(\ln{x})\dfrac{1}{x}dx =\displaystyle \int f(\ln{x})d(\ln{x}) f(lnx)x1dx=f(lnx)d(lnx)

  6. ∫ f ( sin ⁡ x ) cos ⁡ x d x = ∫ f ( sin ⁡ x ) d ( sin ⁡ x ) \displaystyle \int f(\sin{x})\cos{x}dx =\displaystyle \int f(\sin{x})d(\sin{x}) f(sinx)cosxdx=f(sinx)d(sinx)

  7. ∫ f ( cos ⁡ x ) sin ⁡ x d x = − ∫ f ( cos ⁡ x ) d ( cos ⁡ x ) \displaystyle \int f(\cos{x})\sin{x}dx =-\displaystyle \int f(\cos{x})d(\cos{x}) f(cosx)sinxdx=f(cosx)d(cosx)

  8. ∫ f ( tan ⁡ x ) 1 cos ⁡ 2 x d x = ∫ f ( tan ⁡ x ) d ( tan ⁡ x ) \displaystyle \int f(\tan{x})\dfrac{1}{\cos^2{x}}dx =\displaystyle \int f(\tan{x})d(\tan{x}) f(tanx)cos2x1dx=f(tanx)d(tanx)

  9. ∫ f ( arcsin ⁡ x ) 1 1 − x 2 = ∫ f ( arcsin ⁡ x ) d ( arcsin ⁡ x ) \displaystyle \int f(\arcsin{x})\dfrac{1}{\sqrt{1-x^2}} =\displaystyle \int f(\arcsin{x})d(\arcsin{x}) f(arcsinx)1x2 1=f(arcsinx)d(arcsinx)

  10. ∫ f ( arctan ⁡ x ) 1 1 + x 2 = ∫ f ( arctan ⁡ x ) d ( arctan ⁡ x ) \displaystyle \int f(\arctan{x})\dfrac{1}{\sqrt{1+x^2}} =\displaystyle \int f(\arctan{x})d(\arctan{x}) f(arctanx)1+x2 1=f(arctanx)d(arctanx)

4.题

(1)第一类换元法的基本应用

  • ∫ 2 c o s 2 x d x \displaystyle \int 2cos2xdx 2cos2xdx
    = ∫ c o s 2 x ( 2 x ) ′ d x = ∫ c o s u d u ∣ u = 2 x = s i n 2 x + C =\displaystyle \int cos2x(2x)'dx=\displaystyle \int cosudu|_{u=2x}=sin2x+C =cos2x(2x)dx=cosuduu=2x=sin2x+C

  • ∫ 2 x e x 2 d x \displaystyle \int 2xe^{x^2}dx 2xex2dx
    = ∫ e x 2 d ( x 2 ) = e x 2 + C =\displaystyle \int e^{x^2}d(x^2)=e^{x^2}+C =ex2d(x2)=ex2+C

  • ∫ 1 3 + 2 x d x \displaystyle \int \dfrac{1}{3+2x}dx 3+2x1dx
    = ∫ 1 2 1 3 + 2 x ( 3 + 2 x ) ′ d x = 1 2 ∫ 1 u d u ∣ u = 3 + 2 x = 1 2 ln ⁡ ∣ 3 + 2 x ∣ + C =\displaystyle \int \dfrac{1}{2}\dfrac{1}{3+2x}(3+2x)'dx=\dfrac{1}{2} \displaystyle \int\dfrac{1}{u}du|_{u=3+2x}=\dfrac{1}{2}\ln|3+2x|+C =213+2x1(3+2x)dx=21u1duu=3+2x=21ln3+2x+C

  • ∫ 1 x ( 1 + 2 ln ⁡ x ) d x \displaystyle \int \dfrac{1}{x(1+2\ln{x})}dx x(1+2lnx)1dx
    = ∫ 1 2 1 1 + 2 ln ⁡ x d ( 1 + 2 ln ⁡ x ) = 1 2 ln ⁡ ( 1 + 2 ln ⁡ x ) + C =\displaystyle \int \dfrac{1}{2} \dfrac{1}{1+2\ln{x}}d(1+2\ln x)=\dfrac{1}{2}\ln{(1+2\ln{x})}+C =211+2lnx1d(1+2lnx)=21ln(1+2lnx)+C

  • ∫ x 1 − x 2 d x \displaystyle \int x \sqrt{1-x^2}dx x1x2 dx
    = ∫ − 1 2 1 − x 2 d ( 1 − x 2 ) = ∫ − 1 2 ( 1 − x 2 ) 1 2 d ( 1 − x 2 ) = − 1 2 2 3 ( 1 − x 2 ) 3 2 + C = − 1 3 ( 1 − x 2 ) 3 2 + C =\displaystyle \int -\frac{1}{2} \sqrt{1-x^2}d(1-x^2)=\displaystyle \int -\frac{1}{2} {(1-x^2)}^{\frac{1}{2}} d(1-x^2)=-\dfrac{1}{2}\dfrac{2}{3}(1-x^2)^{\frac{3}{2}}+C=-\dfrac{1}{3}(1-x^2)^{\frac{3}{2}}+C =211x2 d(1x2)=21(1x2)21d(1x2)=2132(1x2)23+C=31(1x2)23+C

(2)第一类换元法和分式加减分解

  • ∫ x 2 ( x + 2 ) 3 d x \displaystyle \int \dfrac{x^2}{(x+2)^3}dx (x+2)3x2dx
    = ∫ [ ( x + 2 ) − 2 ] 2 ( x + 2 ) 3 d x = ∫ ( u − 2 ) 2 u 3 d u ∣ u = x + 2 = ∫ ( 1 u − 4 u 2 + 4 u 3 ) d u = ln ⁡ ∣ u ∣ + 4 u − 2 u 2 + C = ln ⁡ ∣ x + 2 ∣ + 4 x + 2 − 2 ( x + 2 ) 2 + C =\displaystyle \int \dfrac{[(x+2)-2]^2}{(x+2)^3}dx=\displaystyle \int \dfrac{(u-2)^2}{u^3}du|_{u=x+2}=\displaystyle \int (\dfrac{1}{u} - \dfrac{4}{u^2}+\dfrac{4}{u^3})du=\ln|u|+\dfrac{4}{u}-\dfrac{2}{u^2}+C=\ln|x+2|+\dfrac{4}{x+2}-\dfrac{2}{(x+2)^2}+C =(x+2)3[(x+2)2]2dx=u3(u2)2duu=x+2=(u1u24+u34)du=lnu+u4u22+C=lnx+2+x+24(x+2)22+C

(3)三角函数专题

  • ∫ 1 9 − x 2 d x \displaystyle \int \dfrac{1}{\sqrt{9-x^2}}dx 9x2 1dx
    = ∫ 1 3 1 1 − ( x 3 ) 2 d x = ∫ 1 1 − ( x 3 ) 2 d ( x 3 ) = arcsin ⁡ x 3 + C =\displaystyle \int \dfrac{1}{3} \dfrac{1}{\sqrt{1-(\frac{x}{3})^2}}dx=\displaystyle \int \dfrac{1}{\sqrt{1-(\frac{x}{3})^2}}d(\frac{x}{3})=\arcsin{\frac{x}{3}}+C =311(3x)2 1dx=1(3x)2 1d(3x)=arcsin3x+C

  • ∫ 1 9 + x 2 d x \displaystyle \int \dfrac{1}{9+x^2}dx 9+x21dx
    = ∫ 1 9 1 1 + ( x 3 ) 2 d x = ∫ 1 3 1 1 + ( x 3 ) 2 d ( x 3 ) = 1 3 arctan ⁡ x 3 + C =\displaystyle \int \dfrac{1}{9} \dfrac{1}{1+(\frac{x}{3})^2}dx=\displaystyle \int \dfrac{1}{3} \dfrac{1}{1+(\frac{x}{3})^2}d(\frac{x}{3})=\frac{1}{3}\arctan{\frac{x}{3}}+C =911+(3x)21dx=311+(3x)21d(3x)=31arctan3x+C

二、第二类换元法

1.核心公式

设, x = ϕ ( u ) x=\phi(u) x=ϕ(u)

∫ f ( x ) d x = ∫ f [ ϕ ( u ) ] ϕ ′ ( u ) ( x ) d u ∣ u = ϕ − 1 ( x ) \int f(x)dx=\int f[\phi(u)]\phi'(u)(x)du|_{u=\phi^{-1}(x)} f(x)dx=f[ϕ(u)]ϕ(u)(x)duu=ϕ1(x)

【思路】:

,从整体的角度将其化成基本积分解决。

【用法】:

找到可以化为基本积分形式的 f ( u ) f(u) f(u),然后变换积分部分构造成满足 d u du du的样子。

比如: ∫ 4 x e x 2 d x \displaystyle \int 4xe^{x^2}dx 4xex2dx e x 2 e^{x^2} ex2就是 f ( u ) f(u) f(u),那么其他部分变换成 d u du du的样子, d u = d ( x 2 ) = 2 x d x du=d(x^2)=2xdx du=d(x2)=2xdx,所以 ∫ 4 x e x 2 d x = ∫ 2 e x 2 d ( x 2 ) = 2 e x 2 + C \displaystyle \int 4xe^{x^2}dx=\displaystyle \int 2e^{x^2}d(x^2)=2e^{x^2}+C 4xex2dx=2ex2d(x2)=2ex2+C

2.常用三种变量代换

(1)知识点

  • 被积函数含有 a 2 − x 2 \sqrt{a^2-x^2} a2x2 ,令 x = a sin ⁡ t ( 或 cos ⁡ t ) x=a\sin{t} \quad (或\cos{t}) x=asint(cost) t ∈ ( − π / 2 , π / 2 ) t \in (-\pi/2,\pi/2) t(π/2,π/2)
    高数之不定积分_第1张图片
    sin ⁡ t = x a \sin{t}=\dfrac{x}{a} sint=ax,则第三边为 a 2 − x 2 \sqrt{a^2-x^2} a2x2

  • 被积函数含有 x 2 − a 2 \sqrt{x^2-a^2} x2a2 ,令 x = a sec ⁡ t x=a\sec{t} x=asect t ∈ ( − π / 2 , π / 2 ) t \in (-\pi/2,\pi/2) t(π/2,π/2)
    高数之不定积分_第2张图片
    sec ⁡ t = x a \sec{t}=\dfrac{x}{a} sect=ax,则第三边为 x 2 − a 2 \sqrt{x^2-a^2} x2a2

  • 被积函数含有 a 2 + x 2 \sqrt{a^2+x^2} a2+x2 ,令 x = a tan ⁡ t x=a\tan{t} x=atant t ∈ ( − π / 2 , π / 2 ) t \in (-\pi/2,\pi/2) t(π/2,π/2)
    高数之不定积分_第3张图片
    tan ⁡ t = x a \tan{t}=\dfrac{x}{a} tant=ax,则第三边为 a 2 + x 2 \sqrt{a^2+x^2} a2+x2

(2)题

  • 【代换1】
    ∫ a 2 − x 2 d x ( a > 0 ) \displaystyle \int \sqrt{a^2-x^2}dx \quad (a>0) a2x2 dx(a>0)
    x = a sin ⁡ t x=a\sin{t} x=asint t ∈ ( − π / 2 , π / 2 ) t \in (-\pi/2,\pi/2) t(π/2,π/2),则
    ∫ a 2 − x 2 d x = ∫ a 2 ( 1 − sin ⁡ 2 t ) d x = ∫ a cos ⁡ t d x = ∫ a cos ⁡ t a cos ⁡ t d t = a 2 ∫ 1 + cos ⁡ 2 t 2 d t = a 2 2 ( t + 1 2 sin ⁡ 2 t ) + C = a 2 2 ( t + sin ⁡ t cos ⁡ t ) + C \displaystyle \int \sqrt{a^2-x^2}dx =\displaystyle \int \sqrt{a^2{(1-\sin^2{t})}}dx =\displaystyle \int a\cos{t}dx =\displaystyle \int a\cos{t} a\cos{t}dt =a^2 \displaystyle \int \dfrac{1+\cos{2t}}{2}dt =\dfrac{a^2}{2}(t+\dfrac{1}{2}\sin{2t})+C =\dfrac{a^2}{2}(t+\sin{t}\cos{t})+C a2x2 dx=a2(1sin2t) dx=acostdx=acostacostdt=a221+cos2tdt=2a2(t+21sin2t)+C=2a2(t+sintcost)+C
    sin ⁡ t = x a \sin{t}=\dfrac{x}{a} sint=ax,得 cos ⁡ t = a 2 − x 2 a \cos{t}=\dfrac{\sqrt{a^2-x^2}}{a} cost=aa2x2
    ∫ a 2 − x 2 d x = a 2 2 ( arcsin ⁡ x a + x a a 2 − x 2 a ) + C = a 2 2 arcsin ⁡ x a + x 2 a 2 − x 2 + C \displaystyle \int \sqrt{a^2-x^2}dx =\dfrac{a^2}{2}(\arcsin{\dfrac{x}{a}}+\dfrac{x}{a}\dfrac{\sqrt{a^2-x^2}}{a})+C =\dfrac{a^2}{2}\arcsin{\dfrac{x}{a}}+\dfrac{x}{2}\sqrt{a^2-x^2}+C a2x2 dx=2a2(arcsinax+axaa2x2 )+C=2a2arcsinax+2xa2x2 +C

  • 【代换2】
    ∫ x 2 − a 2 x d x ( a > 0 ) \displaystyle \int \dfrac{\sqrt{x^2-a^2}}{x}dx \quad (a>0) xx2a2 dx(a>0)
    x = a sec ⁡ t x=a\sec{t} x=asect t ∈ ( − π / 2 , π / 2 ) t \in (-\pi/2,\pi/2) t(π/2,π/2),则
    ∫ x 2 − a 2 x d x = ∫ a tan ⁡ t a sec ⁡ t d x = ∫ sin ⁡ t a sin ⁡ t cos ⁡ 2 t d t = a ∫ tan ⁡ 2 t d t = a ∫ ( sec ⁡ 2 t − 1 ) d t = a tan ⁡ t − a t + C \displaystyle \int \dfrac{\sqrt{x^2-a^2}}{x}dx =\displaystyle \int \dfrac{a\tan{t}}{a\sec{t}}dx =\displaystyle \int \sin{t}\dfrac{a\sin{t}}{\cos^2{t}}dt =a\displaystyle \int \tan^2{t}dt =a\displaystyle \int (\sec^2{t}-1)dt =a\tan{t}-at+C xx2a2 dx=asectatantdx=sintcos2tasintdt=atan2tdt=a(sec2t1)dt=atantat+C
    sec ⁡ t = x a \sec{t}=\dfrac{x}{a} sect=ax,有 cos ⁡ t = a x \cos{t}=\dfrac{a}{x} cost=xa,得 t = arccos ⁡ a x t=\arccos{\dfrac{a}{x}} t=arccosxa
    ∫ x 2 − a 2 x d x = a x 2 − a 2 a − a arccos ⁡ a x + C = x 2 − a 2 − a arccos ⁡ a x + C \displaystyle \int \dfrac{\sqrt{x^2-a^2}}{x}dx =\dfrac{a\sqrt{x^2-a^2}}{a}-a\arccos{\dfrac{a}{x}}+C =\sqrt{x^2-a^2}-a\arccos{\dfrac{a}{x}}+C xx2a2 dx=aax2a2 aarccosxa+C=x2a2 aarccosxa+C

  • 【代换3】
    ∫ 1 a 2 + x 2 d x ( a > 0 ) \displaystyle \int \dfrac{1}{\sqrt{a^2+x^2}}dx \quad (a>0) a2+x2 1dx(a>0)
    x = a tan ⁡ t x=a\tan{t} x=atant t ∈ ( − π / 2 , π / 2 ) t \in (-\pi/2,\pi/2) t(π/2,π/2),则
    ∫ 1 a 2 + x 2 d x = ∫ 1 a sec ⁡ t d x = ∫ 1 a sec ⁡ t a sec ⁡ 2 t d t = ∫ sec ⁡ t d t = ln ⁡ ∣ sec ⁡ t + tan ⁡ t ∣ + C \displaystyle \int \dfrac{1}{\sqrt{a^2+x^2}}dx =\displaystyle \int \dfrac{1}{a\sec{t}}dx =\displaystyle \int \dfrac{1}{a\sec{t}}a\sec^2{t}dt =\displaystyle \int \sec{t}dt =\ln{|\sec{t}+\tan{t}|}+C a2+x2 1dx=asect1dx=asect1asec2tdt=sectdt=lnsect+tant+C
    tan ⁡ t = x a \tan{t}=\dfrac{x}{a} tant=ax,得 sec ⁡ t = a 2 + x 2 x \sec{t}=\dfrac{\sqrt{a^2+x^2}}{x} sect=xa2+x2
    ∫ a 2 − x 2 d x = ln ⁡ ∣ a 2 + x 2 x + x a ∣ + C = ln ⁡ ∣ a 2 + x 2 + x ∣ + C − ln ⁡ a = ln ⁡ ∣ a 2 + x 2 + x ∣ + C \displaystyle \int \sqrt{a^2-x^2}dx =\ln{|\dfrac{\sqrt{a^2+x^2}}{x}+\dfrac{x}{a}|}+C =\ln{|{\sqrt{a^2+x^2}}+{x}|}+C-\ln{a} =\ln{|{\sqrt{a^2+x^2}}+{x}|}+C a2x2 dx=lnxa2+x2 +ax+C=lna2+x2 +x+Clna=lna2+x2 +x+C

3.出错点

【dx和dt的转化方向错误】

x = a sin ⁡ t x=a\sin{t} x=asint
正确:对x求导
∫ d x = ∫ x ′ d x = ∫ a cos ⁡ t d t \displaystyle \int dx =\displaystyle \int x'dx =\displaystyle \int a\cos{t}dt dx=xdx=acostdt
错误:对t求导
∫ d x → ∫ t ′ d x = ∫ ( arcsin ⁡ x a ) ′ d t \displaystyle \int dx \to \displaystyle \int t'dx =\displaystyle \int (\arcsin{\dfrac{x}{a}})'dt dxtdx=(arcsinax)dt

4.题

(1)第二类换元法的基本应用

(2)难题

  • ∫ 1 + e x d x \displaystyle \int \sqrt{1+e^x}dx 1+ex dx
    t = 1 + e x t=\sqrt{1+e^x} t=1+ex ,则 x = ln ⁡ ( t 2 − 1 ) x=\ln{(t^2-1)} x=ln(t21)
    ∫ 1 + e x d x = ∫ t d x = ∫ t 1 t 2 − 1 2 t d t = ∫ 2 t 2 t 2 − 1 d t = ∫ 2 ( t 2 − 1 ) + 2 t 2 − 1 d t = ∫ ( 2 + 2 t 2 − 1 ) d t = 2 t + 2 ∫ 1 t 2 − 1 d t = 2 t + 2 ln ⁡ ∣ t − 1 t + 1 ∣ + C = 2 1 + e x + 2 ln ⁡ ∣ 1 + e x − 1 1 + e x + 1 ∣ + C \displaystyle \int \sqrt{1+e^x}dx =\displaystyle \int tdx =\displaystyle \int t\dfrac{1}{t^2-1}2tdt =\displaystyle \int \dfrac{2t^2}{t^2-1}dt =\displaystyle \int \dfrac{2(t^2-1)+2}{t^2-1}dt =\displaystyle \int (2+\dfrac{2}{t^2-1})dt =2t+2\displaystyle \int \dfrac{1}{t^2-1}dt =2t+2\ln|{\dfrac{t-1}{t+1}}|+C =2\sqrt{1+e^x}+2\ln|{\dfrac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}}|+C 1+ex dx=tdx=tt2112tdt=t212t2dt=t212(t21)+2dt=(2+t212)dt=2t+2t211dt=2t+2lnt+1t1+C=21+ex +2ln1+ex +11+ex 1+C

三、分部积分法

1.公式

∫ u d v = u v − ∫ v d u \displaystyle \int udv=uv-\displaystyle \int vdu udv=uvvdu
∫ u v ′ d x = u v − ∫ v u ′ d x \displaystyle \int uv'dx=uv-\displaystyle \int vu'dx uvdx=uvvudx

2.解析

  • ∫ ln ⁡ x ( 1 − x ) 2 d x \displaystyle \int \dfrac{\ln{x}}{(1-x)^2}dx (1x)2lnxdx
    u = ln ⁡ x , v = 1 1 − x ( ∫ ln ⁡ x ( 1 − x ) 2 d x = ∫ ln ⁡ x ( 1 1 − x ) ′ d x ) u=\ln{x},v=\dfrac{1}{1-x} \quad (\displaystyle \int \dfrac{\ln{x}}{(1-x)^2}dx=\displaystyle \int \ln{x} (\dfrac{1}{1-x})'dx) u=lnx,v=1x1((1x)2lnxdx=lnx(1x1)dx)
    = ln ⁡ x 1 − x − ∫ 1 x ( 1 − x ) d x = ln ⁡ x 1 − x − ∫ ( 1 x + 1 1 − x ) d x = ln ⁡ x 1 − x + ln ⁡ ∣ 1 − x ∣ x + C =\dfrac{\ln{x}}{1-x} - \displaystyle \int \dfrac{1}{x(1-x)}dx =\dfrac{\ln{x}}{1-x} - \displaystyle \int (\dfrac{1}{x} +\dfrac{1}{1-x})dx =\dfrac{\ln{x}}{1-x} + \ln{\dfrac{|1-x|}{x}}+C =1xlnxx(1x)1dx=1xlnx(x1+1x1)dx=1xlnx+lnx1x+C

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