原函数: F ′ ( x ) = f ( x ) F'(x)=f(x) F′(x)=f(x)
不定积分: ∫ f ( x ) d x = F ( x ) + C \int f(x)dx=F(x)+C ∫f(x)dx=F(x)+C
不定积分的几何意义:表示一簇积分曲线,这簇积分曲线对应于横坐标 x x x处的切线都相互平行
定理1: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
定理2: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
例1: g ( x ) = sgn x = { − 1 x < 0 0 x = 0 1 x > 0 g(x)=\text{sgn }x=\begin{cases}-1&x<0\\0&x=0\\1&x>0\end{cases} g(x)=sgn x=⎩ ⎨ ⎧−101x<0x=0x>0
设存在原函数 F ( x ) F(x) F(x),根据分段函数
F ( x ) = { − x + C 1 x < 0 x + C 2 x > 0 F(x)=\begin{cases} -x+C_{1} &x<0\\ x+C_{2}&x>0 \end{cases} F(x)={−x+C1x+C2x<0x>0
由于 F ( x ) F(x) F(x)在 x = 0 x=0 x=0处连续,则 C 1 = C 2 = C C_{1}=C_{2}=C C1=C2=C,有
F ( x ) = { − x + C x < 0 x + C x > 0 = ∣ x ∣ + C F(x)=\begin{cases} -x+C&x<0 \\ x+C&x>0 \end{cases}=|x|+C F(x)={−x+Cx+Cx<0x>0=∣x∣+C
根据定义 F ′ ( x ) = g ( x ) F'(x)=g(x) F′(x)=g(x),显然 F ′ ( 0 ) F'(0) F′(0)不存在,因此 g ( x ) g(x) g(x)不存在原函数
即,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 ∫ f ′ ( x ) d x = f ( x ) + C ∫ d f ( x ) = f ( x ) + C ∫ [ f ( x ) ± g ( x ) ] d x = ∫ f ( x ) d x ± ∫ g ( x ) d x \begin{gathered} \begin{aligned} &(\int f(x)dx)'=f(x)\quad &d\int f(x)dx=f(x)dx\\ &\int f'(x)dx=f(x)+C&\int df(x)=f(x)+C \end{aligned}\\ \int[f(x)\pm g(x)]dx=\int f(x)dx\pm\int g(x)dx \end{gathered} (∫f(x)dx)′=f(x)∫f′(x)dx=f(x)+Cd∫f(x)dx=f(x)dx∫df(x)=f(x)+C∫[f(x)±g(x)]dx=∫f(x)dx±∫g(x)dx
∫ 0 d x = C ∫ x α d x = 1 α + 1 x α + 1 + C ∫ 1 x d x = ln ∣ x ∣ + C ∫ a x d x = a x ln a + C ∫ e x d = e x + C ∫ sin x d x = − cos x + C ∫ cos x d x = sin x + C ∫ sec 2 x d x = tan x + C ∫ csc 2 x d x = − cot x + C ∫ 1 cos 2 x d x = tan x + C ∫ 1 sin 2 x d x = − cot x + C ∫ sec x tan x d x = sec x + C ∫ csc x cot x d x = − csc x + C ∫ 1 1 − x 2 d x = arcsin x + C ∫ 1 1 + x 2 d x = arctan x + C ∫ 1 a 2 − x 2 d x = arcsin x a + C ∫ 1 a 2 + x 2 d x = 1 a arctan x a + C ∫ 1 x 2 − a 2 = 1 2 a ln ∣ x − a x + a ∣ + C ∫ 1 x 2 + a 2 d x = ln ( x + x 2 + a 2 ) + C ∫ 1 x 2 − a 2 d x = ln ∣ x + x 2 − a 2 ∣ + C ∫ sec x d x = ln ∣ sec x + tan x ∣ + C ∫ csc x d x = − ln ∣ csc x + cot x ∣ + C \begin{aligned} \int 0dx&=C\\ \int x^{\alpha} dx&=\frac{1}{\alpha+1}x^{\alpha+1}+C\\ \int \frac{1}x dx&=\ln|x|+C\\ \int a^{x}dx&=\frac{a^{x}}{\ln a}+C\\ \int e^{x}d&=e^{x}+C\\ \int \sin xdx&=-\cos x+C\\ \int \cos xdx&=\sin x+C\\ \int \sec^{2}xdx&=\tan x+C\\ \int \csc^{2}xdx&=-\cot x+C\\ \int \frac{1}{\cos^2 x}dx&=\tan x+C\\ \int \frac{1}{\sin^{2}x}dx&=-\cot x+C\\ \int \sec x\tan xdx&=\sec x+C\\ \int \csc x\cot xdx&=-\csc x+C\\ \int \frac{1}{\sqrt{1-x^2}}dx&=\arcsin x+C\\ \int \frac{1}{1+x^2}dx&=\arctan x+C\\ \int \frac{1}{\sqrt{a^{2}-x^{2}}}dx&=\arcsin \frac{x}{a}+C\\ \int \frac{1}{a^{2}+x^{2}}dx&=\frac{1}{a}\arctan \frac{x}{a}+C\\ \int \frac{1}{x^{2}-a^{2}}&= \frac{1}{2a}\ln| \frac{x-a}{x+a}|+C\\ \int \frac{1}{\sqrt{x^{2}+a^{2}}}dx&=\ln(x+\sqrt{x^{2}+a^{2}})+C\\ \int \frac{1}{\sqrt{x^{2}-a^{2}}}dx&=\ln|x+\sqrt{x^{2}-a^{2}}|+C\\ \int \sec xdx&=\ln|\sec x+\tan x|+C\\ \int \csc xdx&=-\ln|\csc x+\cot x|+C \end{aligned} ∫0dx∫xαdx∫x1dx∫axdx∫exd∫sinxdx∫cosxdx∫sec2xdx∫csc2xdx∫cos2x1dx∫sin2x1dx∫secxtanxdx∫cscxcotxdx∫1−x21dx∫1+x21dx∫a2−x21dx∫a2+x21dx∫x2−a21∫x2+a21dx∫x2−a21dx∫secxdx∫cscxdx=C=α+11xα+1+C=ln∣x∣+C=lnaax+C=ex+C=−cosx+C=sinx+C=tanx+C=−cotx+C=tanx+C=−cotx+C=secx+C=−cscx+C=arcsinx+C=arctanx+C=arcsinax+C=a1arctanax+C=2a1ln∣x+ax−a∣+C=ln(x+x2+a2)+C=ln∣x+x2−a2∣+C=ln∣secx+tanx∣+C=−ln∣cscx+cotx∣+C
若 ∫ f ( u ) d u = F ( u ) + C \int f(u)du=F(u)+C ∫f(u)du=F(u)+C
则 ∫ f [ ϕ ( x ) ] ϕ ′ ( x ) d x = ∫ f [ ϕ ( x ) ] d ϕ ( x ) = F [ ϕ ( x ) ] + C \int f[\phi(x)]\phi'(x)dx=\int f[\phi(x)]d \phi(x)=F[\phi(x)]+C ∫f[ϕ(x)]ϕ′(x)dx=∫f[ϕ(x)]dϕ(x)=F[ϕ(x)]+C
例2:计算积分 ∫ d x x ( 4 − x ) = ( ) \int \frac{dx}{\sqrt{x(4-x)}}=() ∫x(4−x)dx=()
原式 = ∫ d ( x − 2 ) 4 − ( x − 2 ) 2 = arcsin x − 2 2 + C 原式 = ∫ d x x 4 − x = 2 ∫ d x 4 − ( x ) 2 = 2 arcsin x 2 + C \begin{aligned} 原式&=\int \frac{d(x-2)}{\sqrt{4-(x-2)^{2}}}=\arcsin \frac{x-2}{2}+C\\ 原式&=\int \frac{dx}{\sqrt{x}\sqrt{4-x}}=2\int \frac{d \sqrt{x}}{\sqrt{4-(\sqrt{x})^{2}}}=2\arcsin \frac{\sqrt{x}}{2}+C \end{aligned} 原式原式=∫4−(x−2)2d(x−2)=arcsin2x−2+C=∫x4−xdx=2∫4−(x)2dx=2arcsin2x+C
例3:计算积分 ∫ 2 − x 3 + 2 x − x 2 d x = ( ) \int \frac{2-x}{\sqrt{3+2x-x^{2}}}dx=() ∫3+2x−x22−xdx=()
分子分母次数相同,分子带根号,考虑整体凑微分
2 − x 3 + 2 x − x 2 d x = ∫ ( 1 − x ) + 1 3 + 2 x − x 2 d x 此处考虑凑 3 + 2 x − x 2 整体的微分 2 − 2 x 又因为倍数不影响 , 所以根据分子的 − x 凑 1 − x = 1 2 ∫ ( 2 − 2 x ) 3 + 2 x − x 2 d x + ∫ d x 3 + 2 x − x 2 = 1 2 ∫ d ( 3 + 2 x − x 2 ) 3 + 2 x − x 2 + ∫ d ( x + 1 ) 4 − ( x − 1 ) 2 = 3 + 2 x − x 2 + arcsin x − 1 2 + C \begin{aligned} \frac{2-x}{\sqrt{3+2x-x^{2}}}dx&=\int \frac{(1-x)+1}{\sqrt{3+2x-x^{2}}}dx\\ &此处考虑凑3+2x-x^{2}整体的微分2-2x\\ &又因为倍数不影响,所以根据分子的-x凑1-x\\ &=\frac{1}{2}\int \frac{(2-2x)}{\sqrt{3+2x-x^{2}}}dx+\int \frac{dx}{\sqrt{3+2x-x^{2}}}\\ &=\frac{1}{2}\int \frac{d(3+2x-x^{2})}{\sqrt{3+2x-x^{2}}}+\int \frac{d(x+1)}{\sqrt{4-(x-1)^{2}}}\\ &=\sqrt{3+2x-x^{2}}+\arcsin \frac{x-1}{2}+C \end{aligned} 3+2x−x22−xdx=∫3+2x−x2(1−x)+1dx此处考虑凑3+2x−x2整体的微分2−2x又因为倍数不影响,所以根据分子的−x凑1−x=21∫3+2x−x2(2−2x)dx+∫3+2x−x2dx=21∫3+2x−x2d(3+2x−x2)+∫4−(x−1)2d(x+1)=3+2x−x2+arcsin2x−1+C
定理3:设 x = ϕ ( t ) x=\phi(t) x=ϕ(t)是单调的、可导的函数,并且 ϕ ′ ( t ) ≠ 0 \phi'(t)\ne0 ϕ′(t)=0。又 ∫ f [ ϕ ( t ) ] ϕ ′ ( t ) d t = F ( t ) + C \int f[\phi(t)]\phi'(t)dt=F(t)+C ∫f[ϕ(t)]ϕ′(t)dt=F(t)+C,则 ∫ f ( x ) d x = ∫ f [ ϕ ( t ) ] ϕ ′ ( t ) d t = F ( t ) + C = F [ ϕ − 1 ( x ) ] + C \int f(x)dx=\int f[\phi(t)]\phi'(t)dt=F(t)+C=F[\phi^{-1}(x)]+C ∫f(x)dx=∫f[ϕ(t)]ϕ′(t)dt=F(t)+C=F[ϕ−1(x)]+C
常用代换
这些选择的范围都是考虑好的,换元完直接开出来就行
例4:求不定积分 ∫ x 2 a 2 − x 2 d x \int \frac{x^{2}}{\sqrt{a^{2}-x^{2}}}dx ∫a2−x2x2dx
令 x = a sin t , t ∈ ( − π 2 , π 2 ) x=a\sin t,t\in(- \frac{\pi}{2},\frac{\pi}{2}) x=asint,t∈(−2π,2π)
选择 t ∈ ( − π 2 , π 2 ) t\in(- \frac{\pi}{2},\frac{\pi}{2}) t∈(−2π,2π)原因
- 能取遍 ( − a , a ) (-a,a) (−a,a)
- a 2 − x 2 = cos 2 t \sqrt{a^{2}-x^{2}}=\sqrt{\cos^{2} t} a2−x2=cos2t,此时 cos t , t ∈ ( − π 2 , π 2 ) \cos t,t\in(- \frac{\pi}{2},\frac{\pi}{2}) cost,t∈(−2π,2π),即 cos t > 0 \cos t>0 cost>0,有 a 2 − x 2 = cos t \sqrt{a^{2}-x^{2}}=\cos t a2−x2=cost
如果选 cos t \cos t cost,范围应该是 t ∈ ( 0 , π ) t\in(0,\pi) t∈(0,π)
∫ x 2 a 2 − x 2 d x = ∫ a 2 sin 2 t a cos t ⋅ a cos t d t = a 2 2 ∫ ( 1 − cos 2 t ) d t = a 2 2 ( t − 1 2 sin 2 t ) + C = a 2 2 arcsin x a − x 2 a 2 − x 2 + C \begin{aligned} \int \frac{x^{2}}{\sqrt{a^{2}-x^{2}}}dx&=\int \frac{a^{2}\sin^{2}t}{a\cos t}\cdot a\cos tdt\\ &=\frac{a^{2}}{2}\int (1-\cos 2t)dt\\ &=\frac{a^{2}}{2}(t- \frac{1}{2}\sin 2t)+C\\ &=\frac{a^{2}}{2}\arcsin \frac{x}{a}- \frac{x}{2}\sqrt{a^{2}-x^{2}}+C \end{aligned} ∫a2−x2x2dx=∫acosta2sin2t⋅acostdt=2a2∫(1−cos2t)dt=2a2(t−21sin2t)+C=2a2arcsinax−2xa2−x2+C
求不定积分 ∫ x 2 + a 2 x 2 d x \int \frac{\sqrt{x^{2}+a^{2}}}{x^{2}}dx ∫x2x2+a2dx
令 x = a tan t , t ∈ ( − π 2 , π 2 ) x=a\tan t,t\in(- \frac{\pi}{2},\frac{\pi}{2}) x=atant,t∈(−2π,2π)
∫ x 2 + a 2 x 2 d x = ∫ a sec t a 2 tan 2 t ⋅ a sec 2 t d t = ∫ 1 sin 2 t cos t d t = ∫ sin 2 t + cos 2 t sin 2 t cos t d t 分母是 sin cos 高次 , 分子为 1 , 常把分子拆开 = ∫ sec t d t + ∫ cos t sin 2 t d t = ln ∣ sec t + tan t ∣ − 1 sin t + C ∫ x 2 + a 2 x 2 d x = ∫ x 2 + a 2 x 2 x 2 + a 2 d x = ∫ d x x 2 + a 2 + ∫ a 2 d x x 3 1 + ( a x ) 2 = ln ( x − x 2 + a 2 ) − 1 2 ∫ d [ 1 + ( a x ) 2 ] 1 + ( a x ) 2 = ln ( x − x 2 + a 2 ) − 1 + ( a x ) 2 + C \begin{aligned} \int \frac{\sqrt{x^{2}+a^{2}}}{x^{2}}dx&=\int \frac{a\sec t}{a^{2}\tan^{2}t}\cdot a\sec^{2}tdt\\ &=\int \frac{1}{\sin^{2}t\cos t}dt\\ &=\int \frac{\sin^{2}t+\cos^{2}t}{\sin^{2}t\cos t}dt\\ &分母是\sin\cos高次,分子为1,常把分子拆开\\ &=\int \sec tdt+\int \frac{\cos t}{\sin^{2}t}dt\\ &=\ln|\sec t+\tan t|- \frac{1}{\sin t}+C\\ \int \frac{\sqrt{x^{2}+a^{2}}}{x^{2}}dx&=\int \frac{x^{2}+a^{2}}{x^{2}\sqrt{x^{2}+a^{2}}}dx\\ &=\int \frac{dx}{\sqrt{x^{2}+a^{2}}}+\int \frac{a^{2}dx}{x^{3}\sqrt{1+(\frac{a}{x})^{2}}}\\ &=\ln(x-\sqrt{x^{2}+a^{2}})- \frac{1}{2}\int \frac{d[1+(\frac{a}{x})^{2}]}{\sqrt{1+(\frac{a}{x})^{2}}}\\ &=\ln(x-\sqrt{x^{2}+a^{2}})-\sqrt{1+(\frac{a}{x})^{2}}+C \end{aligned} ∫x2x2+a2dx∫x2x2+a2dx=∫a2tan2tasect⋅asec2tdt=∫sin2tcost1dt=∫sin2tcostsin2t+cos2tdt分母是sincos高次,分子为1,常把分子拆开=∫sectdt+∫sin2tcostdt=ln∣sect+tant∣−sint1+C=∫x2x2+a2x2+a2dx=∫x2+a2dx+∫x31+(xa)2a2dx=ln(x−x2+a2)−21∫1+(xa)2d[1+(xa)2]=ln(x−x2+a2)−1+(xa)2+C
求不定积分 ∫ x 2 − a 2 x d x \int \frac{\sqrt{x^{2}-a^{2}}}{x}dx ∫xx2−a2dx
令 x = a sec t , t ∈ ( 0 , π 2 ) x=a\sec t,t\in(0,\frac{\pi}{2}) x=asect,t∈(0,2π)
∫ x 2 − a 2 x d x = ∫ a tan t a sec t ⋅ a sec t tan t d t = a ∫ tan 2 t d t = a ∫ ( sec 2 t − 1 ) d t = a ( tan t − t ) + C = x 2 − a 2 − a arccos a x + C \begin{aligned} \int \frac{\sqrt{x^{2}-a^{2}}}{x}dx&=\int \frac{a\tan t}{a\sec t}\cdot a\sec t\tan tdt\\ &=a\int \tan^{2}tdt\\ &=a\int(\sec^{2}t-1)dt\\ &=a(\tan t-t)+C\\ &=\sqrt{x^{2}-a^{2}}-a\arccos \frac{a}{x}+C \end{aligned} ∫xx2−a2dx=∫asectatant⋅asecttantdt=a∫tan2tdt=a∫(sec2t−1)dt=a(tant−t)+C=x2−a2−aarccosxa+C
例5:求不定积分 ∫ 1 + e x d x \int \sqrt{1+e^{x}}dx ∫1+exdx
令 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(t2−1)
∫ 1 + e x d x = ∫ 2 t 2 t 2 − 1 d x = 2 ∫ ( 1 + 1 t 2 − 1 ) d t = 2 t + ln ∣ t − 1 t + 1 ∣ + C = 2 1 + e 2 + ln 1 + e x − 1 1 + e x + 1 + C \begin{aligned} \int \sqrt{1+e^{x}}dx&=\int \frac{2t^{2}}{t^{2}-1}dx\\ &=2\int(1+ \frac{1}{t^{2}-1})dt\\ &=2t+\ln| \frac{t-1}{t+1}|+C\\ &=2\sqrt{1+e^{2}}+\ln \frac{\sqrt{1+e^{x}}-1}{\sqrt{1+e^{x}}+1}+C \end{aligned} ∫1+exdx=∫t2−12t2dx=2∫(1+t2−11)dt=2t+ln∣t+1t−1∣+C=21+e2+ln1+ex+11+ex−1+C
∫ u d v = u v − ∫ v d u \int udv=uv-\int vdu ∫udv=uv−∫vdu
适用于两类不同函数相乘
按照”反对幂指三“或”反对幂三指“的顺序(记住一个就行,建议前者),谁靠后谁先凑微分(凑 v v v)
即
例6:计算 ∫ ln x ( 1 − x ) 2 d x \int \frac{\ln x}{(1-x)^{2}}dx ∫(1−x)2lnxdx
∫ ln x ( 1 − x ) 2 d x = ∫ ln x d 1 1 − x = ln x 1 − x − ∫ d x x ( 1 − x ) = ln x 1 − x − ∫ ( 1 x + 1 1 − x ) d x = ln x 1 − x + ln ∣ 1 − x ∣ x + C \begin{aligned} \int \frac{\ln x}{(1-x)^{2}}dx&=\int \ln x d \frac{1}{1-x}\\ &=\frac{\ln x}{1-x}-\int \frac{dx}{x(1-x)}\\ &=\frac{\ln x}{1-x}-\int(\frac{1}{x}+ \frac{1}{1-x})dx\\ &=\frac{\ln x}{1-x}+\ln \frac{|1-x|}{x}+C \end{aligned} ∫(1−x)2lnxdx=∫lnxd1−x1=1−xlnx−∫x(1−x)dx=1−xlnx−∫(x1+1−x1)dx=1−xlnx+lnx∣1−x∣+C
常见积不出的积分
∫ e x 2 d x , ∫ sin x x d x , ∫ cos x x d x \int e^{x^{2}}dx,\int \frac{\sin x}{x}dx,\int \frac{\cos x}{x}dx ∫ex2dx,∫xsinxdx,∫xcosxdx
例7: ∫ x + 5 x 2 − 6 x + 13 d x = ( ) \int \frac{x+5}{x^{2}-6x+13}dx=() ∫x2−6x+13x+5dx=()
分子次数少于分母一次,分子整体凑分母微分
分子次数少于分母二次,分母凑平方公式
分子次数多于等于分母,约分分离常数和多项式
∫ x + 5 x 2 − 6 x + 13 d x = 1 2 ∫ d ( x 2 − 6 x + 13 ) x 2 − 6 x + 13 ⏟ 少于分母一次 + 8 ∫ d ( x − 3 ) ( x − 3 ) 2 + 2 2 ⏟ 少于分母二次 = 1 2 ln ( x 2 − 6 x + 13 ) + 4 arctan x − 3 2 + C \begin{aligned} \int \frac{x+5}{x^{2}-6x+13}dx&= \frac{1}{2}\underbrace{\int \frac{d(x^{2}-6x+13)}{x^{2}-6x+13}}_{少于分母一次}+8\underbrace{\int \frac{d(x-3)}{(x-3)^{2}+2^{2}}}_{少于分母二次}\\ &=\frac{1}{2}\ln(x^{2}-6x+13)+4\arctan \frac{x-3}{2}+C \end{aligned} ∫x2−6x+13x+5dx=21少于分母一次 ∫x2−6x+13d(x2−6x+13)+8少于分母二次 ∫(x−3)2+22d(x−3)=21ln(x2−6x+13)+4arctan2x−3+C
∫ d x x ( x 9 + 1 ) = ( ) \int \frac{dx}{x(x^{9}+1)}=() ∫x(x9+1)dx=()
∫ d x x ( x 9 + 1 ) = 1 9 ∫ d x 9 x 9 ( x 9 + 1 ) = 1 9 ln ∣ x 9 x 9 + 1 ∣ + C 凑微分降幂 ∫ d x x ( x 9 + 1 ) = ∫ ( 1 + x 9 ) − x 9 x ( x 9 + 1 ) d x 加项减项拆 = ∫ d x x − ∫ x 8 x 9 + 1 d x = ln ∣ x ∣ − 1 9 ln ( x 9 + 1 ) + C ∫ d x x ( x 9 + 1 ) = ∫ d x x 10 ( 1 + 1 x 9 ) = 1 9 ∫ d ( 1 + 1 x 9 ) 1 + 1 x 9 = − 1 9 ln ∣ 1 + 1 x 9 ∣ + C \begin{aligned} \int \frac{dx}{x(x^{9}+1)}&=\frac{1}{9}\int \frac{dx^{9}}{x^{9}(x^{9}+1)}=\frac{1}{9}\ln| \frac{x^{9}}{x^{9}+1}|+C\\ &凑微分降幂\\ \int \frac{dx}{x(x^{9}+1)}&=\int \frac{(1+x^{9})-x^{9}}{x(x^{9}+1)}dx \\ &加项减项拆\\ &=\int \frac{dx}{x}-\int \frac{x^{8}}{x^{9}+1}dx\\ &=\ln|x|- \frac{1}{9}\ln(x^{9}+1)+C\\ \int \frac{dx}{x(x^{9}+1)}&=\int \frac{dx}{x^{10}(1+ \frac{1}{x^{9}})}\\ &=\frac{1}{9}\int \frac{d(1+\frac{1}{x^{9}})}{1+\frac{1}{x^{9}}}\\ &=- \frac{1}{9}\ln|1+ \frac{1}{x^{9}}|+C \end{aligned} ∫x(x9+1)dx∫x(x9+1)dx∫x(x9+1)dx=91∫x9(x9+1)dx9=91ln∣x9+1x9∣+C凑微分降幂=∫x(x9+1)(1+x9)−x9dx加项减项拆=∫xdx−∫x9+1x8dx=ln∣x∣−91ln(x9+1)+C=∫x10(1+x91)dx=91∫1+x91d(1+x91)=−91ln∣1+x91∣+C
关于加项减项拆
对于分式 P ( x ) Q ( x ) \frac{P(x)}{Q(x)} Q(x)P(x),如果
Q ( x ) = b 0 ( x − a ) α ⋯ ( x − b ) β ( x 2 + p x + q ) λ ⋯ ( x 2 + r x + s ) μ Q(x)=b_{0}(x-a)^{\alpha}\cdots (x-b)^{\beta}(x^{2}+px+q)^{\lambda}\cdots(x^{2}+rx+s)^{\mu} Q(x)=b0(x−a)α⋯(x−b)β(x2+px+q)λ⋯(x2+rx+s)μ
其中 p 2 − 4 q < 0 , ⋯ . r 2 − 4 s < 0 p^{2}-4q<0,\cdots.r^{2}-4s<0 p2−4q<0,⋯.r2−4s<0
则
P ( x ) Q ( x ) = A 1 ( x − a ) α + A 2 ( x − a ) α − 1 + ⋯ + A α x − a + ⋯ + B 1 ( x − b ) β + B 2 ( x − b ) β − 1 + ⋯ + B β x − b + ⋯ + M 1 x + N 1 ( x 2 + p x + q ) λ + M 2 x + N 2 ( x 2 + p x + q ) λ − 1 + ⋯ + M λ x + N λ x 2 + p x + q + ⋯ + R 1 x + S 1 ( x 2 + r x + s ) μ + R 2 x + S 2 ( x 2 + r x + s ) μ − 1 + ⋯ + R μ x + S μ x 2 + r x + s \begin{aligned} \frac{P(x)}{Q(x)}&= \frac{A_{1}}{(x-a)^{\alpha}}+ \frac{A_{2}}{(x-a)^{\alpha-1}}+\cdots+ \frac{A_{\alpha}}{x-a}+\cdots+ \frac{B_{1}}{(x-b)^{\beta}}+\frac{B_{2}}{(x-b)^{\beta-1}}+\cdots\\ &+\frac{B_{\beta}}{x-b}+\cdots+ \frac{M_{1}x+N_{1}}{(x^{2}+px+q)^{\lambda}}+\frac{M_{2}x+N_{2}}{(x^{2}+px+q)^{\lambda-1}}+\cdots+\frac{M_{\lambda}x+N_{\lambda}}{x^{2}+px+q}+\\ &\cdots+\frac{R_{1}x+S_{1}}{(x^{2}+rx+s)^{\mu}}+\frac{R_{2}x+S_{2}}{(x^{2}+rx+s)^{\mu-1}}+\cdots+\frac{R_{\mu}x+S_{\mu}}{x^{2}+rx+s} \end{aligned} Q(x)P(x)=(x−a)αA1+(x−a)α−1A2+⋯+x−aAα+⋯+(x−b)βB1+(x−b)β−1B2+⋯+x−bBβ+⋯+(x2+px+q)λM1x+N1+(x2+px+q)λ−1M2x+N2+⋯+x2+px+qMλx+Nλ+⋯+(x2+rx+s)μR1x+S1+(x2+rx+s)μ−1R2x+S2+⋯+x2+rx+sRμx+Sμ
其中 A 1 , ⋯ , A α , B 1 , ⋯ , B β , M 1 , ⋯ , M λ , N 1 , ⋯ , N λ , R 1 , ⋯ , R μ , S 1 , ⋯ , S μ A_{1},\cdots,A_{\alpha},B_{1},\cdots,B_{\beta},M_{1},\cdots,M_{\lambda},N_{1},\cdots,N_{\lambda},R_{1},\cdots,R_{\mu},S_{1},\cdots,S_{\mu} A1,⋯,Aα,B1,⋯,Bβ,M1,⋯,Mλ,N1,⋯,Nλ,R1,⋯,Rμ,S1,⋯,Sμ等都是常数
在上述有理分式分解中,应注意到以下两点
一般方法(万能代换):令 tan x 2 = t \tan \frac{x}{2}=t tan2x=t
∫ R ( sin x , cos x ) d x = ∫ R ( 2 t 1 + t 2 , 1 − t 2 1 + t 2 ) 2 1 + t 2 d t \int R(\sin x,\cos x)dx=\int R \left(\frac{2t}{1+t^{2}},\frac{1-t^{2}}{1+t^{2}}\right) \frac{2}{1+t^{2}}dt ∫R(sinx,cosx)dx=∫R(1+t22t,1+t21−t2)1+t22dt
特殊方法(三角变形,换元,分部)
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
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
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
注意要提前凑好 d u du du
例8:求 ∫ d x 1 + sin x \int \frac{dx}{1+\sin x} ∫1+sinxdx
分母出现 1 − sin x , 1 − cos x 1-\sin x,1-\cos x 1−sinx,1−cosx一般化成 sin 2 x , cos 2 x \sin^{2}x,\cos^{2}x sin2x,cos2x
原式 = ∫ 1 − sin x cos 2 x d x = tan x − 1 cos x + C \begin{aligned} 原式&=\int \frac{1-\sin x}{\cos^{2}x}dx=\tan \frac{x-1}{\cos x}+C \end{aligned} 原式=∫cos2x1−sinxdx=tancosxx−1+C
或利用万能代换
令 tan x 2 = t \tan \frac{x}{2}=t tan2x=t,则
原式 = ∫ 1 1 + 2 t 1 + t 2 ⋅ 2 1 + t 2 d t = ∫ 2 d t ( 1 + t ) 2 = − 2 1 + t + C = − 2 1 + tan x 2 + C \begin{aligned} 原式&=\int \frac{1}{1+\frac{2t}{1+t^{2}}}\cdot \frac{2}{1+t^{2}}dt\\ &=\int \frac{2dt}{(1+t)^{2}}\\ &=- \frac{2}{1+t}+C\\ &=- \frac{2}{1+\tan \frac{x}{2}}+C \end{aligned} 原式=∫1+1+t22t1⋅1+t22dt=∫(1+t)22dt=−1+t2+C=−1+tan2x2+C
一般尤其高次三角函数不考虑万能代换
求 ∫ d x sin ( 2 x ) + 2 sin x = ( ) \int \frac{dx}{\sin(2x)+2\sin x}=() ∫sin(2x)+2sinxdx=()
原式 = ∫ d x 2 sin x ( cos x + 1 ) 发现满足 R ( − sin x , cos x ) = − R ( sin x , cos x ) 为了凑 d u = d cos x , 上下同乘 sin x = ∫ sin x d x 2 ( 1 − cos 2 x ) ( 1 + cos x ) = cos x = u − 1 2 ∫ d u ( 1 − u ) ( 1 + u ) 2 = − 1 2 ∫ [ A 1 − u + B ( 1 + u ) 2 + C 1 + u ] d u 通分有 A u 2 + 2 A u + A + B − B u + C − C u 2 = 1 即 { A − C = 0 2 A − B = 0 A + B + C = 1 解得 { A = 1 4 B = 1 2 C = 1 4 = − 1 4 ∫ ( 1 2 1 1 − u + 1 ( 1 + u ) 2 + 1 2 1 1 + u ) d u = 1 8 [ ln 1 − u 1 + u + 2 1 + u ] + C = 1 8 ln 1 − cos x 1 + cos x + 1 4 ( 1 + cos x ) + C \begin{aligned} 原式&=\int \frac{dx}{2\sin x(\cos x+1)}\\ &发现满足R(-\sin x,\cos x)=-R(\sin x,\cos x)\\ &为了凑du=d\cos x,上下同乘\sin x\\ &=\int \frac{\sin xdx}{2(1-\cos^{2}x)(1+\cos x)}\\ &\overset{\cos x=u}{=}- \frac{1}{2}\int \frac{du}{(1-u)(1+u)^{2}}\\ &=- \frac{1}{2}\int[\frac{A}{1-u}+ \frac{B}{(1+u)^{2}}+ \frac{C}{1+u}]du\\ &通分有Au^{2}+2Au+A+B-Bu+C-Cu^{2}=1\\ &即\begin{cases} A-C=0 \\ 2A-B=0 \\ A+B+C=1 \end{cases}解得\begin{cases} A=\frac{1}{4} \\ B=\frac{1}{2} \\ C=\frac{1}{4} \end{cases}\\ &=- \frac{1}{4}\int(\frac{1}{2} \frac{1}{1-u}+ \frac{1}{(1+u)^{2}}+ \frac{1}{2} \frac{1}{1+u})du\\ &=\frac{1}{8}[\ln \frac{1-u}{1+u}+ \frac{2}{1+u}]+C\\ &=\frac{1}{8}\ln \frac{1-\cos x}{1+\cos x}+ \frac{1}{4(1+\cos x)}+C \end{aligned} 原式=∫2sinx(cosx+1)dx发现满足R(−sinx,cosx)=−R(sinx,cosx)为了凑du=dcosx,上下同乘sinx=∫2(1−cos2x)(1+cosx)sinxdx=cosx=u−21∫(1−u)(1+u)2du=−21∫[1−uA+(1+u)2B+1+uC]du通分有Au2+2Au+A+B−Bu+C−Cu2=1即⎩ ⎨ ⎧A−C=02A−B=0A+B+C=1解得⎩ ⎨ ⎧A=41B=21C=41=−41∫(211−u1+(1+u)21+211+u1)du=81[ln1+u1−u+1+u2]+C=81ln1+cosx1−cosx+4(1+cosx)1+C
令 a x + b c x + d n = t \sqrt[n]{\frac{ax+b}{cx+d}}=t ncx+dax+b=t
例9:计算 ∫ 1 x x + 1 x d x \int \frac{1}{x}\sqrt{\frac{x+1}{x}}dx ∫x1xx+1dx
令 x + 1 x \sqrt{\frac{x+1}{x}} xx+1,则 x = 1 t 2 − 1 , d x = − 2 t ( t 2 − 1 ) 2 d t x=\frac{1}{t^{2}-1},dx=- \frac{2t}{(t^{2}-1)^{2}}dt x=t2−11,dx=−(t2−1)22tdt
∫ 1 x x + 1 x d x = ∫ ( t 2 − 1 ) t − 2 t ( t 2 − 1 ) 2 d t = − 2 ∫ ( 1 + 1 t 2 − 1 ) d t = − 2 ( t + 1 2 ln ∣ t − 1 t + 1 ∣ ) + C \begin{aligned} \int \frac{1}{x}\sqrt{\frac{x+1}{x}}dx&=\int (t^{2}-1)t \frac{-2t}{(t^{2}-1)^{2}}dt\\ &=-2\int(1+ \frac{1}{t^{2}-1})dt\\ &=-2(t+ \frac{1}{2}\ln| \frac{t-1}{t+1}|)+C \end{aligned} ∫x1xx+1dx=∫(t2−1)t(t2−1)2−2tdt=−2∫(1+t2−11)dt=−2(t+21ln∣t+1t−1∣)+C
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