前段时间复习完了高数第四章的内容，我参考《复习全书·基础篇》和老师讲课的内容对这一章的知识点进行了整理，形成了这篇笔记，方便在移动设备上进行访问和后续的补充修改。
f ( x ) f(x) f(x)的原函数的全体成为 f ( x ) f(x) f(x)的不定积分，记为 ∫ f ( x ) d x \int f(x)dx ∫f(x)dx.
如果 F ( x ) F(x) F(x)为 f ( x ) f(x) f(x)的一个原函数，则有
∫ f ( x ) d x = F ( x ) + C (4.1) \int {f(x)}dx = F(x) + C \tag{4.1} ∫f(x)dx=F(x)+C(4.1)
其中 C C C为任意常数
若 f ( x ) f(x) f(x)在区间 I I I上连续，则 f ( x ) f(x) f(x)在区间 I I I上一定存在原函数
证明不存在的定理若 f ( x ) f(x) f(x)在区间 I I I上有第一类间断点，则 f ( x ) f(x) f(x)在区间 I I I上没有原函数
( ∫ f ( x ) d x ) ′ = f ( x ) , d ( ∫ f ( x ) d x ) = f ( x ) d x (4.2) (\int {f(x)}dx)' = f(x), d (\int {f(x)}dx) = f(x)dx \tag{4.2} (∫f(x)dx)′=f(x),d(∫f(x)dx)=f(x)dx(4.2)
∫ f ′ ( x ) d x = f ( x ) + C , ∫ d f ( x ) d x = f ( x ) + C (4.3) \int {f'(x)}dx = f(x) + C, \int d{f(x)}dx = f(x) + C \tag{4.3} ∫f′(x)dx=f(x)+C,∫df(x)dx=f(x)+C(4.3)
∫ f ( x ) ± g ( x ) d x = ∫ f ( x ) d x ± ∫ g ( x ) d x (4.4) \int {f(x) \pm g(x)}dx = \int {f(x)}dx \pm \int {g(x)}dx \tag{4.4} ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx(4.4)
∫ k f ( x ) d x = k ∫ f ( x ) d x , ( k = C ) (4.5) \int k{f(x)}dx = k \int {f(x)}dx, (k = C) \tag{4.5} ∫kf(x)dx=k∫f(x)dx,(k=C)(4.5)
∫ 0 d x = C (4.6) \int {0}dx = C \tag{4.6} ∫0dx=C(4.6)
∫ x a d x = 1 a + 1 x α + 1 + C , ( α ≠ − 1 ) (4.7) \int {x^a}dx = \frac{1}{a+1}x^{\alpha + 1} + C, (\alpha \ne -1) \tag{4.7} ∫xadx=a+11xα+1+C,(α=−1)(4.7)
∫ 1 x d x = ln ∣ x ∣ + C (4.8) \int \frac{1}{x} dx = \ln|x| + C \tag{4.8} ∫x1dx=ln∣x∣+C(4.8)
∫ a x d x = a x ln a + C , ( a > 0 , a ≠ 1 ) (4.9) \int a^x dx = \frac{a^x}{\ln a} + C, (a > 0, a \ne 1) \tag{4.9} ∫axdx=lnaax+C,(a>0,a=1)(4.9)
∫ e x d x = e x + C (4.10) \int e^x dx = e^x + C \tag{4.10} ∫exdx=ex+C(4.10)
∫ sin x d x = − cos ( x ) + C (4.11) \int \sin x dx = - \cos(x) + C \tag{4.11} ∫sinxdx=−cos(x)+C(4.11)
∫ cos ( x ) d x = sin ( x ) + C (4.12) \int \cos(x) dx = \sin(x) + C \tag{4.12} ∫cos(x)dx=sin(x)+C(4.12)
∫ sec 2 x d x = tan ( x ) + C (4.13) \int \sec^2 x dx = \tan(x) + C \tag{4.13} ∫sec2xdx=tan(x)+C(4.13)
∫ csc 2 x d x = − ctg x + C (4.14) \int \csc^2 x dx = -\ctg x + C \tag{4.14} ∫csc2xdx=−ctgx+C(4.14)
∫ sec x tan x d x = sec x + C (4.15) \int \sec x \tan x dx = \sec x + C \tag{4.15} ∫secxtanxdx=secx+C(4.15)
∫ csc x ctg x d x = − csc x + C (4.16) \int \csc x \ctg x dx = - \csc x + C \tag{4.16} ∫cscxctgxdx=−cscx+C(4.16)
∫ 1 1 − x 2 d x = arcsin x + C (4.17) \int \frac{1}{\sqrt{1 - x^2}}dx = \arcsin x + C \tag{4.17} ∫1−x2 1dx=arcsinx+C(4.17)
证明4.17： 凑微分法∫ 1 1 − x 2 d x = ∫ d x a 1 − ( x a ) 2 d x = ∫ d ( x a ) 1 − ( x a ) 2 d x = arcsin x + C { \begin{aligned} \int \frac{1}{\sqrt{1 - x^2}}dx &= \int \dfrac{dx}{a \sqrt{1 - (\dfrac{x}{a}})^2} dx \\ &= \int \frac{d (\dfrac{x}{a})}{\sqrt{1 - (\dfrac{x}{a}})^2} dx \\ &= \arcsin x + C \end{aligned} } ∫1−x2 1dx=∫a1−(ax )2dxdx=∫1−(ax )2d(ax)dx=arcsinx+C
∫ 1 1 + x 2 d x = arctan x + C (4.18) \int \frac{1}{{1 + x^2}}dx = \arctan x + C \tag{4.18} ∫1+x21dx=arctanx+C(4.18)
∫ 1 a 2 + x 2 d x = 1 a arctan x a + C (4.19) \int \frac{1}{{a^2 + x^2}}dx = \frac{1}{a} \arctan \frac{x}{a} + C \tag{4.19} ∫a2+x21dx=a1arctanax+C(4.19)
∫ 1 x 2 − a 2 d x = 1 2 a ln ∣ x − a x + a ∣ + C (4.20) \int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \ln|\frac{x-a}{x+a}| + C \tag{4.20} ∫x2−a21dx=2a1ln∣x+ax−a∣+C(4.20)
∫ 1 x 2 + a 2 d x = ln ( x + x 2 + a 2 ) + C (4.21) \int \frac{1}{\sqrt{x^2 + a^2}} dx = \ln (x + \sqrt{x^2 + a^2}) + C \tag{4.21} ∫x2+a2 1dx=ln(x+x2+a2 )+C(4.21)
证明4.21： 第二类换元法，令 x = a tan t x = a\tan t x=atant∫ 1 x 2 + a 2 d x = ∫ a sec 2 t a sec t d t = ∫ sec t d t = ln ∣ sec t + tan t ∣ + C = ln ∣ x + x 2 + a 2 ∣ − ln a + C = ln ∣ x + x 2 + a 2 ∣ + C { \begin{aligned} \int \frac{1}{\sqrt{x^2 + a^2}} dx &= \int \frac{a\sec^2 t}{a \sec t} dt = \int \sec t dt \\ &= \ln |\sec t + \tan t| + C \\ &= \ln |x + \sqrt{x^2 + a^2}| - \ln a+ C\\ &= \ln |x + \sqrt{x^2 + a^2}| + C \end{aligned} } ∫x2+a2 1dx=∫asectasec2tdt=∫sectdt=ln∣sect+tant∣+C=ln∣x+x2+a2 ∣−lna+C=ln∣x+x2+a2 ∣+C
∫ 1 x 2 − a 2 d x = ln ∣ x + x 2 − a 2 ∣ + C (4.22) \int \frac{1}{\sqrt{x^2 - a^2}} dx = \ln |x + \sqrt{x^2 - a^2}| + C \tag{4.22} ∫x2−a2 1dx=ln∣x+x2−a2 ∣+C(4.22)
证明4.22： 第二类换元法，令 x = a sec t x = a\sec t x=asect∫ 1 x 2 − a 2 d x = ∫ a sec t tan t a tan t d t = ∫ sec t d t = ln ∣ sec t + tan t ∣ + C = ln ∣ x + x 2 − a 2 ∣ − ln a + C = ln ∣ x + x 2 − a 2 ∣ + C { \begin{aligned} \int \frac{1}{\sqrt{x^2 - a^2}} dx &= \int \frac{a\sec t \tan t}{a \tan t} dt = \int \sec t dt \\ &= \ln |\sec t + \tan t| + C \\ &= \ln |x + \sqrt{x^2 - a^2}| - \ln a+ C\\ &= \ln |x + \sqrt{x^2 - a^2}| + C \end{aligned} } ∫x2−a2 1dx=∫atantasecttantdt=∫sectdt=ln∣sect+tant∣+C=ln∣x+x2−a2 ∣−lna+C=ln∣x+x2−a2 ∣+C
∫ sec x d x = ln ∣ sec x + tan x ∣ + C (4.23) \int {\sec x} dx = \ln |\sec x + \tan x| + C \tag{4.23} ∫secxdx=ln∣secx+tanx∣+C(4.23)
证明4.23： 凑微分法∫ sec x d x = ∫ sec x [ sec x + tan x ] sec x + tan x d x = ∫ sec 2 x + sec x tan x sec x + tan x d x = ∫ d ( sec x + tan x ) sec x + tan x = ln ∣ sec x + tan x ∣ + C { \begin{aligned} \int {\sec x} dx &= \int \frac{\sec x[\sec x + \tan x]}{\sec x + \tan x} dx &=& \int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} dx \\ &= \int \frac{d(\sec x + \tan x)}{\sec x + \tan x}\\ &= \ln |\sec x + \tan x| + C \end{aligned} } ∫secxdx=∫secx+tanxsecx[secx+tanx]dx=∫secx+tanxd(secx+tanx)=ln∣secx+tanx∣+C=∫secx+tanxsec2x+secxtanxdx
∫ csc x d x = − ln ∣ csc x + ctg x ∣ + C (4.24) \int {\csc x} dx = -\ln |\csc x + \ctg x| + C \tag{4.24} ∫cscxdx=−ln∣cscx+ctgx∣+C(4.24)
证明4.24： 凑微分法∫ csc x d x = ∫ csc x [ csc x + ctg x ] csc x + ctg x d x = ∫ csc 2 x + csc x ctg x csc x + ctg x d x = ∫ d ( csc x + ctg x ) csc x + ctg x = ln ∣ csc x + ctg x ∣ + C { \begin{aligned} \int {\csc x} dx &= \int \frac{\csc x[\csc x + \ctg x]}{\csc x + \ctg x} dx &=& \int \frac{\csc^2 x + \csc x \ctg x}{\csc x + \ctg x} dx \\ &= \int \frac{d(\csc x + \ctg x)}{\csc x + \ctg x}\\ &= \ln |\csc x + \ctg x| + C \end{aligned} } ∫cscxdx=∫cscx+ctgxcscx[cscx+ctgx]dx=∫cscx+ctgxd(cscx+ctgx)=ln∣cscx+ctgx∣+C=∫cscx+ctgxcsc2x+cscxctgxdx
∫ f [ φ ( x ) ] φ ′ ( x ) d x = ∫ f [ φ ( x ) ] d φ x = F ( φ ( x ) ) + C (4.25) \int f[\varphi(x)]\varphi '(x) dx = \int f[\varphi(x)] d\varphi x = F(\varphi(x)) + C \tag{4.25} ∫f[φ(x)]φ′(x)dx=∫f[φ(x)]dφx=F(φ(x))+C(4.25)
∫ f [ φ ( t ) ] φ ′ ( t ) d t = F ( φ ( t ) ) + C \int f[\varphi(t)]\varphi '(t) dt = F(\varphi(t)) + C ∫f[φ(t)]φ′(t)dt=F(φ(t))+C
则
∫ f ( x ) d x = ∫ f [ φ ( t ) ] φ ′ ( t ) d t = F ( φ ( t ) ) + C = F [ φ − 1 ( x ) ] + C (4.26) \int {f(x)} dx = \int f[\varphi(t)]\varphi '(t) dt = F(\varphi(t)) + C = F[\varphi^{-1}(x)] + C \tag{4.26} ∫f(x)dx=∫f[φ(t)]φ′(t)dt=F(φ(t))+C=F[φ−1(x)]+C(4.26)
注：式中对 φ ( t ) \varphi (t) φ(t)求导的部分容易被遗漏
常用的三种变量代换被积函数含有 a 2 − x 2 \sqrt{a^2 - x^2} a2−x2 ，令 x = a sin x x = a\sin x x=asinx（或 a cos x a \cos x acosx).
被积函数含有 x 2 + a 2 \sqrt{x^2 + a^2} x2+a2 ，令 x = a tan x x = a\tan x x=atanx.
被积函数含有 x 2 − a 2 \sqrt{x^2 - a^2} x2−a2 ，令 x = a sec x x = a\sec x x=asecx.
∫ u d v = u v − ∫ v d u (4.27) \int u dv = uv - \int v du \tag{4.27} ∫udv=uv−∫vdu(4.27)
分部积分法中 u , v u,v u,v的选取 把多项式以外的函数凑进微分号，因为对多项式求导若干次后能够将其化为常数项∫ p n ( x ) e α x d x , ∫ p n ( x ) sin α x d x , ∫ p n ( x ) cos α x d x \int p_n(x)e^{\alpha x} dx, \int p_n(x)\sin \alpha x dx, \int p_n(x)\cos \alpha x dx ∫pn(x)eαxdx,∫pn(x)sinαxdx,∫pn(x)cosαxdx
把指数函数或三角函数凑进微分号都可以，但把指数凑进去更简单∫ e α x sin β x d x , ∫ e α x cos β x \int e^{\alpha x}\sin \beta x dx, \int e^{\alpha x}\cos \beta x ∫eαxsinβxdx,∫eαxcosβx
把多项式凑进微分号，多项式以外的函数方便求导，不方便积分∫ p n ( x ) ln x d x , ∫ p n ( x ) arctan x d x , ∫ p n ( x ) arcsin x d x \int p_n(x)\ln x dx, \int p_n(x)\arctan x dx, \int p_n(x)\arcsin x dx ∫pn(x)lnxdx,∫pn(x)arctanxdx,∫pn(x)arcsinxdx
∫ R ( sin x , cos x ) d x = ∫ R ( 2 t 1 + t 2 , 1 − t 2 1 + t 2 ) d t (4.28) \int R(\sin x, \cos x) dx = \int R(\frac{2t}{1 + t^2}, \frac{1 - t^2}{1 + t^2}) dt \tag{4.28} ∫R(sinx,cosx)dx=∫R(1+t22t,1+t21−t2)dt(4.28)
特殊方法（三角变形，换元，分解） 几种常用的换元法 若 R ( − sin x , cos x ) = − R ( sin x , cos x ) R(- \sin x, \cos x) = - R(\sin x, \cos x) R(−sinx,cosx)=−R(sinx,cosx)，则令 u = cos x u = \cos x u=cosx，或凑 d cos x d\cos x dcosx.若 R ( sin x , − cos x ) = − R ( sin x , cos x ) R(\sin x, - \cos x) = - R(\sin x, \cos x) R(sinx,−cosx)=−R(sinx,cosx)，则令 u = sin x u = \sin x u=sinx，或凑 d sin x d\sin x dsinx.若 R ( − sin x , − cos x ) = R ( sin x , cos x ) R(- \sin x, - \cos x) = R(\sin x, \cos x) R(−sinx,−cosx)=R(sinx,cosx)，则令 u = tan x u = \tan x u=tanx，或凑 d tan x d\tan x dtanx.令 a x + b c x + d n = t \sqrt[n]{\dfrac{ax + b}{cx + d}} = t ncx+dax+b =t，将其转化为有理函数积分进行计算
两个概念
原函数不定积分三种方法
第一类换元法第二类换元法分部积分法三种形式
有理函数三角有理式简单无理函数