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不定积分的计算
2024-11-10
Notes
Math

常用积分表#

必背#

(1)∫kdx=kx+C(2)∫xadx=xa+1a+1+C(3)∫1xdx=ln⁡∣x∣+C(4)∫11+x2dx=arctan⁡x+C(5)∫11−x2dx=arcsin⁡x+C(6)∫axdx=1ln⁡aax+C(7)∫cos⁡xdx=sin⁡x+C(8)∫sin⁡xdx=−cos⁡x+C(9)∫sec⁡2xdx=tan⁡x+C(10)∫csc⁡2xdx=−cot⁡x+C(11)∫sec⁡xtan⁡xdx=sec⁡x+C(12)∫csc⁡xcot⁡xdx=−csc⁡x+C(13)∫sec⁡xdx=ln⁡∣sec⁡x+tan⁡x∣+C(14)∫csc⁡xdx=ln⁡∣csc⁡x−cot⁡x∣+C\begin{array}{} (1) \int kdx = kx +C & (2) \int x^adx = \frac{x^{a+1}}{a+1}+C\\ (3) \int{\frac{1}{x}}dx = \ln|x|+C& (4) \int \frac{1}{1+x^2}dx = \arctan x+C \\ (5)\int{\frac{1}{\sqrt{ 1-x^2 }}dx}=\arcsin x+C & (6)\int a^xdx=\frac{1}{\ln a}a^x+C \\ (7)\int \cos xdx = \sin x+C & (8)\int \sin xdx= -\cos x+C \\ (9)\int \sec^2xdx= \tan x+C & (10)\int \csc^2xdx=-\cot x+C \\ (11)\int \sec x\tan xdx=\sec x+C&(12)\int \csc x\cot xdx=-\csc x+C \\ (13)\int \sec xdx=\ln|\sec x+\tan x|+C&(14)\int \csc xdx=\ln|\csc x-\cot x|+C \end{array}

考前必背#

(1) ∫tan⁡x dx=−ln⁡∣cos⁡x∣+C(2) ∫cot⁡x dx=ln⁡∣sin⁡x∣+C(3) ∫dxx2+a2=1aarctan⁡ ⁣(xa)+C(4) ∫dxx2±a2=ln⁡ ⁣∣ x+x2±a2 ∣+C(5) ∫dxa2−x2=arcsin⁡ ⁣(xa)+C(6) ∫dxa2−x2=12aln⁡ ⁣∣ a+x  a−x ∣+C\begin{array}{ll} (1)\,\int \tan x\,dx = -\ln|\cos x| + C & (2)\,\int \cot x\,dx = \ln|\sin x| + C \\ (3)\,\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan\!\Bigl(\frac{x}{a}\Bigr) + C & (4)\,\int \frac{dx}{\sqrt{x^2 \pm a^2}} = \ln\!\bigl|\,x + \sqrt{x^2 \pm a^2}\,\bigr| + C \\ (5)\,\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\!\Bigl(\frac{x}{a}\Bigr) + C & (6)\,\int \frac{dx}{a^2 - x^2} = \frac{1}{2a}\ln\!\Bigl|\frac{\,a + x\,}{\,a - x\,}\Bigr| + C \end{array}

例题#

例一 ∫1x4(x2+1)dx\int\frac{{1}}{x^4(x^2+1)}dx#

[!NOTE] 关键点 通过分项的方法: 抄分母, 加一个, 减一个.

原式=∫(1+x2)−x2x4(x2+1)dx=∫1x4dx−∫1x2(x2+1)dx=∫1x4dx−∫1+x2−x2x2(x2+1)dx=∫1x4dx−∫1x2dx+∫1x2+1dx=−13x−3−1x+arctan⁡x+C\begin{aligned} \text{原式}=\int\frac{{(1+x^2)-x^2}}{x^4(x^2+1)}dx&=\int \frac{1}{x^4}dx-\int \frac{1}{x^2(x^2+1)}dx\\ &=\int\frac{{1}}{x^4}dx-\int{\frac{{1+x^2-x^2}}{x^2(x^2+1)}}dx\\ &=\int \frac{1}{x^4}dx-\int \frac{1}{x^2}dx+\int \frac{1}{x^2+1}dx\\ &=-\frac{1}{3}x^{-3}-\frac{1}{x}+\arctan x+C \end{aligned}

例二 ∫tan⁡2xdx\int \tan^2xdx#

[!NOTE] 关键点 tan⁡2x+1=sec⁡2x\tan^2x+1=\sec^2x 可以由 sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1 两边同时除以 cos⁡2x\cos^2x 得到

原式=∫(sec⁡2−1)dx=tan⁡x−x+C\begin{align} \text{原式} &=\int(\sec^2-1)dx\\ \\ &=\tan x-x+C \end{align}

例三 ∫cos⁡2x2dx\int \cos^2{\frac{x}{2}}dx#

[!NOTE] 关键点:三角恒等变换 sin⁡αcos⁡α=12sin⁡2α\sin\alpha \cos\alpha=\frac{1}{2}\sin2\alpha cos⁡2α=1+cos⁡2α2\cos^2\alpha=\frac{{1+\cos 2\alpha}}{2} sin⁡2α=1−cos⁡2α2\sin^2\alpha=\frac{{1-\cos2\alpha}}{2}

原式=∫1+cos⁡x2dx=∫12dx+∫cos⁡x2dx=12x+12sin⁡x+C\begin{align} \text{原式}&=\int{\frac{{1+\cos x}}{2}}dx \\ &=\int \frac{1}{2}dx+\int \frac{{\cos x}}{2}dx \\ &=\frac{1}{2}x+\frac{1}{2}\sin x+C \end{align}

常用凑微分公式#

∫f(ax+b) dx=1a∫f(ax+b) d(ax+b)(a≠0)∫f(sin⁡x)cos⁡x dx=∫f(sin⁡x) d(sin⁡x)∫f(ln⁡x)1x dx=∫f(ln⁡x) d(ln⁡x)∫f(x)dxx=2∫f(x) d(x)∫f(arctan⁡x)dx1+x2=∫f(arctan⁡x) d(arctan⁡x)\begin{array}{l} \int f(ax+b)\,dx = \frac{1}{a}\int f(ax+b)\,d(ax+b)\quad (a\neq0)\\[6pt] \int f(\sin x)\cos x\,dx = \int f(\sin x)\,d(\sin x)\\[6pt] \int f(\ln x)\frac{1}{x}\,dx = \int f(\ln x)\,d(\ln x)\\[6pt] \int f\bigl(\sqrt{x}\bigr)\frac{dx}{\sqrt{x}} = 2\int f\bigl(\sqrt{x}\bigr)\,d\bigl(\sqrt{x}\bigr)\\[6pt] \int f(\arctan x)\frac{dx}{1+x^2} = \int f(\arctan x)\,d(\arctan x) \end{array}

例题#

例四 ∫cos⁡(ex)exdx\int \cos (e^x)e^xdx#

[!NOTE] 关键点: 第一类换元积分法 ∫f[φ(x)]φ′(x) dx=(∫f(u) du)u=φ(x)\int f \left[ \varphi (x) \right] \varphi' (x) \, dx = \left ( \int f (u) \, du \right)_{u = \varphi (x)}

df(x)=f′(x)dx→dex=exdxdf(x)=f'(x)dx\to de^x=e^xdx 本例中可将 exe^x 看作 uu

原式=∫cos⁡exdex=sin⁡ex+C\begin{align} \text{原式}&=\int \cos e^xde^x \\ &=\sin e^x +C \end{align}

例五 ∫tan⁡xdx\int \tan xdx#

原式=∫sin⁡xcos⁡xdx=−∫1cos⁡xdcos⁡x=−ln⁡∣cos⁡x∣+C\begin{align} \text{原式}&=\int \frac{{\sin x}}{\cos x}dx \\ &=-\int \frac{1}{\cos x}d\cos x \\ &=-\ln|\cos x|+C \end{align}

例六 ∫1a2−x2dx\int \frac{{1}}{\sqrt{ a^2-x^2 }}dx#

原式=∫1a2(1−(xa)2)dx=∫11−(xa)2⋅1adx=∫11+(xa)2dxa=arcsin⁡xa+C\begin{align} \text{原式}&=\int{\frac{{1}}{\sqrt{ a^2\left( 1-\left( \frac{x}{a} \right)^2 \right) }}}dx \\ &=\int {\frac{{1}}{\sqrt{ 1-\left( \frac{x}{a} \right)^2 }}}{\cdot\frac{{1}}{a}}dx \\ &=\int {\frac{{1}}{\sqrt{ 1+\left( \frac{x}{a} \right)^2 }}}d{\frac{x}{a}} \\ &=\arcsin{\frac{x}{a}}+C \end{align}

例七 ∫dxx(1+x)\int \frac{{dx}}{\sqrt{ x }(1+x)}#

[!NOTE] 关键点 dx=(x)′dx=12xdxd\sqrt{ x }=(\sqrt{ x })'dx=\frac{{1}}{2\sqrt{ x }}dx 所以 dxx=2dx{\frac{dx}{\sqrt{ x }}=2d\sqrt{ x }}

原式=2∫11+(x)2dx=2arctan⁡x\begin{align} \text{原式}&=2\int \frac{{1}}{1+(\sqrt{ x })^2}d\sqrt{ x } \\ &=2\arctan \sqrt{ x } \end{align}

常见换元法#

[!NOTE] 第二类换元积分法 ∫f(x) dx=(∫f[φ(t)]φ′(t) dt)t=φ−1(x)\int f (x) \, dx = \left ( \int f \left[ \varphi (t) \right] \varphi'(t) \, dt \right)_{t = \varphi^{-1}(x)}

三角代换#

被积函数含 a2−x2(a>0)\sqrt{ a^2-x^2 }(a>0), 令 x=asin⁡t,t∈[−π2,π2]x=a\sin t,t \in\left[ {-\frac{\pi}{2},{\frac{\pi}{2}}} \right], 则 a2−x2=acos⁡t\sqrt{ a^2-x^2 }=a\cos t. 被积函数含 x2+a2(a>0)\sqrt{ x^2+a^2 }(a>0), 令 x=atan⁡t,t∈[−π2,π2]x=a\tan t,t \in\left[ {-\frac{\pi}{2},{\frac{\pi}{2}}} \right], 则 x2+a2=asec⁡t\sqrt{ x^2+a^2 }=a\sec t. 被积函数含 x2−a2(a>0)\sqrt{ x^2-a^2 }(a>0), 令 x=asec⁡t,t∈[0,π2)∪(π2,π]x=a\sec t,t \in\left[ {0},{\frac{\pi}{2}})\cup({\frac{\pi}{2},\pi} \right], 则 x2−a2=atan⁡t\sqrt{ x^2-a^2 }=a\tan t.

根式代换#

见例八

例题#

例八 ∫dx1+x\int \frac{dx}{{1+\sqrt{ x }}}#

[!NOTE] 关键点: 根式代换

令t=x原式=∫dt21+t=∫2tdt1+t=2∫(t+1)−1t+1dt=2[∫dt−∫11+tdt]=2[t−ln⁡(t+1)]+C=2x⋅2ln⁡(x+1)+C\begin{align} \text{令}t=\sqrt{ x } \\ \text{原式}&=\int {\frac{{dt^2}}{1+t}} \\ &=\int {\frac{{2tdt}}{1+t}}=2\int {\frac{{(t+1)-1}}{t+1}}dt \\ &=2\left[ \int dt-\int \frac{1}{1+t}dt \right] \\ &=2[t-\ln (t+1)]+C \\ &=2\sqrt{ x }\cdot 2\ln(\sqrt{ x }+1)+C \end{align}

例九 ∫a2−x2dx\int \sqrt{ a^2-x^2 }dx#

[!NOTE] 关键点: 三角代换 利用图形将新元替换为旧元

令x=asin⁡t原式=∫acos⁡t⋅acos⁡tdt=∫a2cos⁡2tdt=a2∫cos⁡2tdt=a22∫(1+cos⁡2t)dt=a22(∫dt+12∫cos⁡2td2t)=a22t+a24sin⁡2t+C\begin{align} \text{令}x=a\sin t \\ \text{原式}&=\int a\cos t\cdot a\cos tdt \\ &=\int a^2\cos^2tdt=a^2\int \cos^2tdt=\frac{a^2}{2}\int(1+\cos 2t)dt \\ &=\frac{a^2}{2}\left( \int dt + \frac{1}{2}\int \cos 2t d 2t\right) \\ &=\frac{a^2}{2}t+\frac{a^2}{4}\sin 2t+C \end{align}

image.png 所以

原式=a22arcsin⁡xa+a22⋅xa⋅a2−x2a+C=a22arcsin⁡xa+xaa2−x2+C\begin{align} \text{原式}&=\frac{a^2}{2}\arcsin \frac{{x}}{a} +\frac{a^2}{2}\cdot \frac{x}{a}\cdot{\frac{\sqrt{ a^2-x^2 }}{a}}+C \\ &=\frac{a^2}{2}\arcsin \frac{x}{a}+\frac{x}{a}\sqrt{ a^2-x^2 }+C \end{align}

分部积分法#

∫u(x)dv(x)=u(x)v(x)−∫v(x)du(x)\begin{align} \int u(x)dv(x)=u(x)v(x)-\int v(x)du(x) \end{align}

[!NOTE] 技巧 一般可依次选择 u 的顺序为: 反三角函数, 对数函数, 幂函数, 指数函数. 三角函数 (反对幂指三)

例题#

例十 ∫x2ln⁡xdx\int x^2\ln xdx#

[!NOTE] 技巧 灵活结合换元法

原式=∫ln⁡xx2dx=13∫ln⁡xdx3=13(ln⁡x⋅x3−∫x3dln⁡x)=13(x3ln⁡x−∫x2dx)=13(x3ln⁡x−13x3)+C\begin{align} \text{原式}&=\int \ln x x^2dx=\frac{1}{3}\int \ln x dx^3 \\ &=\frac{1}{3}\left( \ln x\cdot x^3-\int x^3d\ln x \right) \\ &=\frac{1}{3}\left( x^3\ln x-\int x^2dx \right) \\ &=\frac{1}{3}\left( x^3\ln x-\frac{1}{3}x^3 \right)+C \end{align}

例十一 ∫xex(x+1)2dx\int {\frac{{xe^x}}{(x+1)^2}}dx#

[!NOTE] 关键 ∫ex(x+1)2dx=∫ex(x+1)2d(x+1)\int \frac{e^x}{(x+1)^2}dx=\int \frac{e^x}{(x+1)^2}d(x+1)

原式=∫(x+1)−1(x+1)2⋅exdx=∫exx+1dx−∫ex(x+1)2dx=∫exx+1dx+∫exd1x+1=∫exx+1dx+ex⋅1x+1−∫exx+1dx=exx+1+C\begin{align} \text{原式}&=\int {\frac{{(x+1)-1}}{(x+1)^2}}\cdot e^xdx=\int \frac{e^x}{x+1}dx-\int \frac{e^x}{(x+1)^2}dx \\ &=\int \frac{e^x}{x+1}dx+\int e^xd \frac{{1}}{x+1} \\ &=\int \frac{e^x}{x+1}dx+e^x\cdot \frac{{1}}{x+1}-\int \frac{e^x}{x+1}dx \\ &=\frac{e^x}{x+1}+C \end{align}

含 sin⁡mx⋅cos⁡nx\sin^mx\cdot \cos^nx 的积分#

sin⁡x奇次，cos⁡x偶次→d(cos⁡x) 或 d(sin⁡x)cos⁡x奇次，sin⁡x偶次→d(sin⁡x) 或 d(cos⁡x)cos⁡x和sin⁡x同为偶次或奇次→d(tan⁡x) 或 d(cos⁡x)\begin{align} \sin x \text{奇次，}\cos x \text{偶次} \rightarrow d(\cos x) \text{ 或 } d(\sin x)\\ \cos x \text{奇次，}\sin x \text{偶次} \rightarrow d(\sin x) \text{ 或 } d(\cos x)\\ \cos x \text{和}\sin x \text{同为偶次或奇次} \rightarrow d(\tan x) \text{ 或 } d(\cos x) \end{align}

例题#

例十二 ∫1sin⁡3x⋅cos⁡xdx\int \frac{1}{\sin^3x\cdot \cos x}dx#

[!NOTE] 关键点 将 sec⁡4x\sec^4x 替换为 sec⁡2x⋅sec⁡2x\sec^2x\cdot \sec^2x 用后面的 sec⁡2x\sec^2x 与 dxdx 凑微分

上下同时除以cos⁡4x原式=∫sec⁡4xtan⁡3xdx=∫(tan⁡2x+1)dtan⁡xtan⁡3x=∫(1tan⁡x+tan⁡−3x)dtan⁡x=ln⁡∣tan⁡x∣−12tan⁡−2x+C\begin{align} \text{上下同时除以}\cos^4x \\ \text{原式}&=\int \frac{\sec^4x}{\tan^3x}dx \\ &=\int {\frac{{(\tan^2x+1)d\tan x}}{\tan^3x}}=\int \left( {\frac{1}{\tan x}+\tan^{-3}x} \right)d\tan x \\ &=\ln|\tan x|-\frac{1}{2}\tan^{-2}x+C \end{align}
不定积分的计算
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发布于
2024-11-10
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