math_基本导数公式@积分公式@部分推导@数学工具@公式识别@计算推演工具

文章目录

导数积分公式表🎈

积分计算器

  • 积分计算@图像

  • Wolfram|Alpha: Computational Intelligence (wolframalpha.com)

    • Wolfram|Alpha Examples: Calculus & Analysis (wolframalpha.com)

    • 该网站提供了解题步骤,(需要开会员版)

    • 例:计算 ∫ − ∞ x 1 1 + exp ⁡ ( − x ) d x \int_{-\infin}^{x}\frac{1}{1+\exp(-x)}dx ∫−∞x​1+exp(−x)1​dx

      • https://www.wolframalpha.com/input?i=int+1%2F%281%2Be%5E%28-x%29%29+x%3D-infinity..x

      • 答案是：

      • ∫ − ∞ x 1 1 + exp ⁡ ( − t ) d t = ∫ − ∞ x exp ⁡ ( t ) exp ⁡ ( t ) + 1 d t = ∫ − ∞ x 1 exp ⁡ ( t ) + 1 d ( exp ⁡ ( t ) ) = ∫ − ∞ x 1 exp ⁡ ( t ) + 1 d ( exp ⁡ ( t ) + 1 ) 记 u = exp ⁡ ( t ) + 1 , u ∈ ( 1 , exp ⁡ ( x ) + 1 ) = ∫ 1 exp ⁡ ( x ) + 1 1 u d u = ln ⁡ ∣ u ∣ ∣ 1 exp ⁡ ( x ) + 1 = ln ⁡ ( exp ⁡ ( x ) + 1 ) − ln ⁡ ( 1 ) = ln ⁡ ( exp ⁡ ( x ) + 1 ) \int_{-\infin}^{x}\frac{1}{1+\exp(-t)}dt =\int_{-\infin}^{x}\frac{\exp(t)}{\exp(t)+1}dt \\ =\int_{-\infin}^{x}\frac{1}{\exp(t)+1}\mathrm{d}({\exp(t)}) \\ =\int_{-\infin}^{x}\frac{1}{\exp(t)+1}\mathrm{d}({\exp(t)}+1) \\ 记u=\exp(t)+1,u\in(1,\exp(x)+1) \\ =\int_{1}^{\exp(x)+1}\frac{1}{u}\mathrm{d}{u} \\ =\left.\ln|u|\right|_{1}^{\exp(x)+1} =\ln(\exp(x)+1)-\ln(1) =\ln(\exp(x)+1) ∫−∞x​1+exp(−t)1​dt=∫−∞x​exp(t)+1exp(t)​dt=∫−∞x​exp(t)+11​d(exp(t))=∫−∞x​exp(t)+11​d(exp(t)+1)记u=exp(t)+1,u∈(1,exp(x)+1)=∫1exp(x)+1​u1​du=ln∣u∣∣1exp(x)+1​=ln(exp(x)+1)−ln(1)=ln(exp(x)+1)

数学工具@公式识别@计算推演工具

pictures version

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markdown Table version🎈🎈🎈🎈

  • 渲染不正常的,末尾附带原码,自行用typora渲染

    f ( x ) f(x) f(x) f ′ ( x ) = d d x f ( x ) f'(x)=\frac{d}{\mathrm{d}x}f(x) f′(x)=dxd​f(x) ∫ f ( x ) d x \displaystyle\int{f(x)}\mathrm{d}x ∫f(x)dxNote
    k k k0 k x + C kx+C kx+C
    x 1 2 = x x^{\frac{1}{2}}=\sqrt{x} x21​=x ​ 1 2 x = 1 2 x − 1 2 \frac{1}{2\sqrt{x}}=\frac{1}{2}x^{-\frac{1}{2}} 2x ​1​=21​x−21​ 2 3 x 3 2 + C \frac{2}{3}x^{\frac{3}{2}}+C 32​x23​+C
    1 x = x − 1 \frac{1}{x}=x^{-1} x1​=x−1 − 1 x 2 = − x − 2 -\frac{1}{x^2}=-x^{-2} −x21​=−x−2$\ln{x
    x a x^{a} xa a x a − 1 ax^{a-1} axa−1 1 a + 1 x a + 1 + C \frac{1}{a+1}x^{a+1}+C a+11​xa+1+C a ≠ − 1 a\neq{-1} a=−1
    a x a^x ax a x ln ⁡ a a^x\ln{a} axlna 1 ln ⁡ a a x + C \frac{1}{\ln{a}}a^x+C lna1​ax+C a > 0 , a ≠ 1 a>0, a\neq 1 a>0,a=1
    e x e^x ex e x e^x ex e x + C e^x+C ex+C
    log ⁡ a x \log_a{x} loga​x 1 x ln ⁡ a \frac{1}{x\ln{a}} xlna1​ x log ⁡ a x − 1 ln ⁡ a x + C x\log_{a}x-\frac{1}{\ln{a}}x+C xloga​x−lna1​x+C积分使用分部积分法
    ln ⁡ x \ln{x} lnx 1 x \frac{1}{x} x1​ x ln ⁡ x − x + C x\ln{x}-x+C xlnx−x+C$(\ln
    sin ⁡ x \sin{x} sinx cos ⁡ x \cos{x} cosx − cos ⁡ x + C -\cos{x}+C −cosx+C
    cos ⁡ x \cos{x} cosx − sin ⁡ x -\sin{x} −sinx sin ⁡ x + C \sin{x}+C sinx+C
    sec ⁡ x \sec{x} secx sec ⁡ x tan ⁡ x \sec{x}\tan{x} secxtanx$\ln{\sec{x}+\tan{x}
    csc ⁡ x \csc{x} cscx − csc ⁡ x cot ⁡ x -\csc{x}\cot{x} −cscxcotx$-\ln\csc{x}+\cot{x}
    tan ⁡ x \tan{x} tanx sec ⁡ 2 x \sec^2{x} sec2x$-\ln\cos{x}
    cot ⁡ x \cot{x} cotx − csc ⁡ 2 x -\csc^2{x} −csc2x$\ln\sin{x}
    sin ⁡ 2 x \sin^2{x} sin2x x 2 − sin ⁡ 2 x 4 + C \frac{x}{2}-\frac{\sin{2x}}{4}+C 2x​−4sin2x​+C sin ⁡ 2 x = 1 2 ( 1 − cos ⁡ 2 x ) \sin^2{x}=\frac{1}{2}(1-\cos{2x}) sin2x=21​(1−cos2x)
    cos ⁡ 2 x \cos^2{x} cos2x x 2 + sin ⁡ 2 x 4 + C \frac{x}{2}+\frac{\sin{2x}}{4}+C 2x​+4sin2x​+C cos ⁡ 2 x = 1 2 ( 1 + cos ⁡ 2 x ) \cos^2{x}=\frac{1}{2}(1+\cos{2x}) cos2x=21​(1+cos2x)
    sec ⁡ 2 x \sec^2{x} sec2x tan ⁡ x + C \tan{x}+C tanx+C
    csc ⁡ 2 x \csc^2{x} csc2x − cot ⁡ x + C -\cot{x}+C −cotx+C
    tan ⁡ 2 x \tan^2{x} tan2x tan ⁡ x − x + C \tan{x}-x+C tanx−x+C tan ⁡ 2 x = sec ⁡ 2 x − 1 \tan^2{x}=\sec^2x-1 tan2x=sec2x−1
    cot ⁡ 2 x \cot^2{x} cot2x − cot ⁡ x − x + C -\cot{x}-x+C −cotx−x+C cot ⁡ 2 x = csc ⁡ 2 x − 1 \cot^2{x}=\csc^2{x}-1 cot2x=csc2x−1
    sec ⁡ x tan ⁡ x \sec{x}\tan{x} secxtanx sec ⁡ x + C \sec{x}+C secx+C
    csc ⁡ x cot ⁡ x \csc{x}\cot{x} cscxcotx − csc ⁡ x + C -\csc{x}+C −cscx+C
    arcsin ⁡ x \arcsin{x} arcsinx 1 1 − x 2 \frac{1}{\sqrt{1-x^2}} 1−x2 ​1​
    arccos ⁡ x \arccos{x} arccosx − 1 1 − x 2 -\frac{1}{\sqrt{1-x^2}} −1−x2 ​1​
    arctan ⁡ x \arctan{x} arctanx 1 1 + x 2 \frac{1}{1+x^2} 1+x21​
    a r c cot ⁡ x \mathrm{arc}\cot{x} arccotx − 1 1 + x 2 -\frac{1}{1+x^2} −1+x21​
    1 x 2 ± a 2 \frac{1}{\sqrt{x^2\pm a^2}} x2±a2 ​1​$\ln{x}+\sqrt{x^2\pm a^2}
    1 a 2 − x 2 \frac{1}{\sqrt{a^2-x^2}} a2−x2 ​1​ arcsin ⁡ x a + C \arcsin{\frac{x}{a}}+C arcsinax​+C
    1 1 − x 2 \frac{1}{\sqrt{1-x^2}} 1−x2 ​1​ arcsin ⁡ x + C \arcsin{x}+C arcsinx+C
    1 x 2 + a 2 \frac{1}{x^2+a^2} x2+a21​ 1 a arctan ⁡ x a + C \frac{1}{a}\arctan{\frac{x}{a}}+C a1​arctanax​+C
    1 1 + x 2 \frac{1}{1+x^2} 1+x21​ arctan ⁡ x + C \arctan{x}+C arctanx+C
    1 x 2 − a 2 \frac{1}{x^2-a^2} x2−a21​$\frac{1}{2a}\ln\frac{x-a}{x+a}
    a 2 − x 2 \sqrt{a^2-x^2} a2−x2 ​ 1 2 ( a 2 arcsin ⁡ x a + x a 2 − x 2 ) + C \frac{1}{2}(a^2\arcsin{\frac{x}{a}}+x\sqrt{a^2-x^2})+C 21​(a2arcsinax​+xa2−x2 ​)+C$(a>
    x 2 ± a 2 \sqrt{x^2\pm{}a^2} x2±a2 ​$\frac{1}{2}(x\sqrt{x2\pm{}a2}\pm{}a^2\lnx+\sqrt{x2\pm{}a2}

Notes

  • NoteA:

    • csc ⁡ x − cot ⁡ x = ( csc ⁡ x + cot ⁡ x ) − 1 1 sin ⁡ x − cos ⁡ x sin ⁡ x = 1 1 sin ⁡ x + cos ⁡ x sin ⁡ x 1 − cos ⁡ x sin ⁡ x = sin ⁡ x 1 + cos ⁡ x 即 1 − cos ⁡ 2 x = sin ⁡ 2 x , 这显然成立 \csc{x}-\cot{x}=(\csc{x}+\cot{x})^{-1} \\ \frac{1}{\sin{x}}-\frac{\cos{x}}{\sin{x}}=\frac{1}{\frac{1}{\sin{x}}+\frac{\cos{x}}{\sin{x}}} \\ \frac{1-\cos{x}}{\sin{x}}=\frac{\sin{x}}{1+\cos{x}} 即 \\1-\cos^2{x}=\sin^2{x},这显然成立 cscx−cotx=(cscx+cotx)−1sinx1​−sinxcosx​=sinx1​+sinxcosx​1​sinx1−cosx​=1+cosxsinx​即1−cos2x=sin2x,这显然成立
补充

  • ( ln ⁡ ∣ x ∣ ) ′ = 1 x ( x ≠ 0 ) (\ln|x|)'=\frac{1}{x}(x\neq{0}) (ln∣x∣)′=x1​(x=0)

    • 分段分析,去掉绝对值

      • { ln ⁡ x , x > 0 ln ⁡ ( − x ) , x < 0 分别求导 : { 1 x , x > 0 1 − x ( − 1 ) = 1 x , x < 0 ∴ ( ln ⁡ ∣ x ∣ ) ′ = 1 x , x ≠ 0 \\ \begin{cases} \ln{x},x>0 \\ \ln{(-x)},x<0 \end{cases} \\ 分别求导: \begin{cases} \frac{1}{x},x>0 \\ \frac{1}{-x}{(-1)}=\frac{1}{x},x<0 \end{cases} \\ \therefore (\ln|x|)'=\frac{1}{x},x\neq{0} {lnx,x>0ln(−x),x<0​分别求导:{x1​,x>0−x1​(−1)=x1​,x<0​∴(ln∣x∣)′=x1​,x=0

几个积分公式的推导和记忆方法🎈

x 2 ± a 2 x^2\pm{a^2} x2±a2

  1. ∫ 1 x 2 ± a 2 d x = ln ⁡ ∣ x + x 2 ± a 2 ∣ + C \int \frac{1}{\sqrt{x^2\pm a^2}}dx=\ln|x+\sqrt{x^2\pm a^2}|+C ∫x2±a2 ​1​dx=ln∣x+x2±a2 ​∣+C

    1. p = p ± = x 2 ± A = x 2 ± a 2 ; A = a 2 p=p_{\pm}=\sqrt{x^2\pm A}=\sqrt{x^2\pm a^2};A=a^2 p=p±​=x2±A ​=x2±a2 ​;A=a2
    2. ∫ 1 x 2 − a 2 d x = ∫ 1 p d x = ln ⁡ ∣ x + p ∣ + C ★ \int \frac{1}{\sqrt{x^2-a^2}}dx=\int \frac{1}{p}dx=\ln|x+p|+C\bigstar ∫x2−a2 ​1​dx=∫p1​dx=ln∣x+p∣+C★
    3. 可由三角换元推导
    4. 形如表达式 ∫ 1 x 2 + A \int \frac{1}{\sqrt{x^2+A}} ∫x2+A ​1​的形式出现,就可以套用本公式
a 2 − x 2 \sqrt{a^2-{x^2}} a2−x2 ​

  • ∫ a 2 − x 2 d x = a 2 2 a r c s i n x a + 1 2 x a 2 − x 2 + C = 1 2 ( a 2 a r c s i n x a + x a 2 − x 2 ) + C \int \sqrt{a^2-x^2}dx=\frac{a^2}{2}arcsin{\frac{x}{a}}+\frac{1}{2}x\sqrt{a^2-x^2}+C=\frac{1}{2}(a^2arcsin{\frac{x}{a}}+x\sqrt{a^2-x^2})+C ∫a2−x2 ​dx=2a2​arcsinax​+21​xa2−x2 ​+C=21​(a2arcsinax​+xa2−x2 ​)+C

    • p = a 2 − x 2 p=\sqrt{a^2-x^2} p=a2−x2 ​

    • ∫ p d x = 1 2 a 2 ∫ 1 p d x + 1 2 x p = 1 2 ( a 2 ∫ 1 p d x + x p ) + C \int p dx=\frac{1}{2}a^2\int \frac{1}{p}dx+\frac{1}{2}xp=\frac{1}{2}(a^2\int \frac{1}{p}dx+xp)+C ∫pdx=21​a2∫p1​dx+21​xp=21​(a2∫p1​dx+xp)+C

x 2 − a 2 \sqrt{x^2-a^2} x2−a2 ​

  • ∫ x 2 − a 2 d x = 1 2 ( x x 2 − a 2 − a 2 ln ⁡ ∣ x + x 2 − a 2 ∣ ) \int \sqrt{x^2-a^2}dx=\frac{1}{2}(x\sqrt{x^2-a^2}-a^2\ln |x+\sqrt{x^2-a^2}|) ∫x2−a2 ​dx=21​(xx2−a2 ​−a2ln∣x+x2−a2 ​∣)

    • 分部积分

      • S = ∫ x 2 − a 2 d x = x x 2 − a 2 − ∫ x d x 2 − a 2 为例方便说明推导和简洁性 , 提前给出如下标记 ( 表达式记号 ) A = x x 2 − a 2 B = ∫ x d x 2 − a 2 P = x 2 − a 2 Q = a 2 ∫ 1 x 2 − a 2 d x = a 2 ln ⁡ ∣ x + x 2 − a 2 ∣ \\ \begin{aligned} S&=\int \sqrt{x^2-a^2}dx \\ &=x\sqrt{x^2-a^2}-\int xd\sqrt{x^2-a^2} \end{aligned} \\为例方便说明推导和简洁性,提前给出如下标记(表达式记号) \\ \begin{aligned} \\A&=x\sqrt{x^2-a^2} \\B&=\int xd\sqrt{x^2-a^2} \\P&=\sqrt{x^2-a^2} \\Q&=a^2\int\frac{1}{\sqrt{x^2-a^2}}dx=a^2\ln |x+\sqrt{x^2-a^2}| \end{aligned} S​=∫x2−a2 ​dx=xx2−a2 ​−∫xdx2−a2 ​​为例方便说明推导和简洁性,提前给出如下标记(表达式记号)ABPQ​=xx2−a2 ​=∫xdx2−a2 ​=x2−a2 ​=a2∫x2−a2 ​1​dx=a2ln∣x+x2−a2 ​∣​

      • B = ∫ x d x 2 − a 2 = ∫ x 2 x 2 − a 2 d x = 分子 + 0 = − a 2 + a 2 ∫ x 2 − a 2 + a 2 x 2 − a 2 d x = ∫ x 2 − a 2 d x + a 2 ∫ 1 x 2 − a 2 d x = S + Q \begin{aligned} B &=\int xd\sqrt{x^2-a^2}\\ &=\int \frac{x^2}{\sqrt{x^2-a^2}}dx \\ &\xlongequal{分子+0=-a^2+a^2}\int \frac{x^2-a^2+a^2}{\sqrt{x^2-a^2}}dx\\ &=\int\sqrt{x^2-a^2}dx+a^2\int\frac{1}{\sqrt{x^2-a^2}}dx\\ \\ &=S+Q \end{aligned} B​=∫xdx2−a2 ​=∫x2−a2 ​x2​dx分子+0=−a2+a2 ∫x2−a2 ​x2−a2+a2​dx=∫x2−a2 ​dx+a2∫x2−a2 ​1​dx=S+Q​

      • S = A − B = A − S − Q 2 S = A − Q → S = 1 2 ( A − Q ) S = 1 2 ( x x 2 − a 2 − a 2 ln ⁡ ∣ x + x 2 − a 2 ∣ ) \\S=A-B=A-S-Q \\2S=A-Q \to S=\frac{1}{2}(A-Q) \\S=\frac{1}{2}(x\sqrt{x^2-a^2}-a^2\ln |x+\sqrt{x^2-a^2}|) S=A−B=A−S−Q2S=A−Q→S=21​(A−Q)S=21​(xx2−a2 ​−a2ln∣x+x2−a2 ​∣)

a 2 + x 2 \sqrt{a^2+x^2} a2+x2 ​

∫ a 2 + x 2 d x = 1 2 ( x a 2 + x 2 + a 2 ln ⁡ ∣ a 2 + x 2 + x ∣ ) + C \displaystyle\int \sqrt{a^2+x^2}dx=\frac{1}{2}(x\sqrt{a^2+x^2}+a^2 \ln|\sqrt{a^2+x^2}+x|)+C ∫a2+x2 ​dx=21​(xa2+x2 ​+a2ln∣a2+x2 ​+x∣)+C

  1. 利用三角换元配合分部积分法可以推导
  2. ∫ s e c 3 t   d t = 1 2 ( s e c x t a n x + ln ⁡ ∣ s e c x + t a n x ∣ ) + C \int sec^3t\ dt=\frac{1}{2}(secxtanx+\ln |secx+tanx|)+C ∫sec3t dt=21​(secxtanx+ln∣secx+tanx∣)+C
  • p = a 2 + x 2 p=\sqrt{a^2+x^2} p=a2+x2 ​
  • A = x p A=xp A=xp
  • Q = a 2 ln ⁡ ∣ x + p ∣ Q=a^2\ln|x+p| Q=a2ln∣x+p∣
  • S = 1 2 ( A + Q ) + C S=\frac{1}{2}(A+Q)+C S=21​(A+Q)+C

幂函数积分的一些常用特值扩充🎃

  • o v e r h e a d : 积分升幂 ( 特例 : 1 x ) ∫ x k d x = 1 k + 1 x k + 1 + C = x p p + C , p = k + 1 ∫ 1 x d x = ∫ x − 1 d x = l n ∣ x ∣ + C ∫ 1 x 2 d x = ∫ x − 2 d x = − x − 1 + C = − 1 x + C ∫ 1 x d x = ∫ x − 1 2 d x = 2 x 1 2 + C = 2 x + C ∫ x d x = ∫ x 1 2 d x = 2 3 x 3 2 + C o v e r h e a d : 求导降幂 ( x k ) ′ = k x k − 1 ( 1 x ) ′ = ( x − 1 ) ′ = − x − 2 = − 1 x 2 ( 1 x 2 ) ′ = ( x − 2 ) ′ = − 2 x − 3 ( 1 x ) ′ = ( x − 1 2 ) ′ = − 1 2 x − 3 2 ( x ) ′ = ( x 1 2 ) ′ = 1 2 x − 1 2 = 1 2 x \\ \begin{aligned} &overhead:积分升幂(特例:\frac{1}{x})\\ &\int x^kdx=\frac{1}{k+1}x^{k+1}+C=\frac{x^p}{p}+C,p=k+1\\ &\int \frac{1}{x}dx=\int x^{-1}dx=ln|x|+C\\ &\int \frac{1}{x^2}dx=\int x^{-2}dx=-x^{-1}+C=-\frac{1}{x}+C\\ &\int \frac{1}{\sqrt{x}}dx=\int x^{-\frac{1}{2}}dx=2x^{\frac{1}{2}}+C=2\sqrt{x}+C\\ &\int \sqrt{x} dx=\int x^{\frac{1}{2}}dx=\frac{2}{3}x^{\frac{3}{2}}+C \end{aligned} \quad \large\begin{aligned} &overhead:求导降幂\\ &(x^k)'=kx^{k-1}\\ &(\frac{1}{x})'=(x^{-1})'=-x^{-2}=-\frac{1}{x^2}\\ &(\frac{1}{x^2})'=(x^{-2})'=-2x^{-3}\\ &(\frac{1}{\sqrt{x}})'=(x^{-\frac{1}{2}})'=-\frac{1}{2}x^{-\frac{3}{2}}\\ &(\sqrt{x})'=(x^{\frac{1}{2}})'=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}} \end{aligned} ​overhead:积分升幂(特例:x1​)∫xkdx=k+11​xk+1+C=pxp​+C,p=k+1∫x1​dx=∫x−1dx=ln∣x∣+C∫x21​dx=∫x−2dx=−x−1+C=−x1​+C∫x ​1​dx=∫x−21​dx=2x21​+C=2x ​+C∫x ​dx=∫x21​dx=32​x23​+C​​overhead:求导降幂(xk)′=kxk−1(x1​)′=(x−1)′=−x−2=−x21​(x21​)′=(x−2)′=−2x−3(x ​1​)′=(x−21​)′=−21​x−23​(x ​)′=(x21​)′=21​x−21​=2x ​1​​

  • ∫ x d x = 1 2 x 2 + c \int xdx=\frac{1}{2}x^2+c ∫xdx=21​x2+c

  • ∫ t a n x d x = ln ⁡ ∣ s e c x ∣ + C \int tanxdx=\ln |secx|+C ∫tanxdx=ln∣secx∣+C

  • ∫ c o t x d x = − ln ⁡ ∣ c s c x ∣ + C = ln ⁡ ∣ s i n x ∣ + C \int cotxdx=- \ln |cscx|+C=\ln |sinx|+C ∫cotxdx=−ln∣cscx∣+C=ln∣sinx∣+C

附:表格原码

  • 由于某些markdown无法很好的从表格中的竖线|区分开来,提供原码,可以用typora等渲染

  		• | $f(x)$                        | $f'(x)=\frac{d}{\mathrm{d}x}f(x)$                 | $\displaystyle\int{f(x)}\mathrm{d}x$                         | Note                                |
    		  | ----------------------------- | ------------------------------------------------- | ------------------------------------------------------------ | ----------------------------------- |
    		  | $k$                           | 0                                                 | $kx+C$                                                       |                                     |
    		  | $x^{\frac{1}{2}}=\sqrt{x}$    | $\frac{1}{2\sqrt{x}}=\frac{1}{2}x^{-\frac{1}{2}}$ | $\frac{2}{3}x^{\frac{3}{2}}+C$                               |                                     |
    		  | $\frac{1}{x}=x^{-1}$          | $-\frac{1}{x^2}=-x^{-2}$                          | $\ln{|x|}+C$                                                 |                                     |
    		  | $x^{a}$                       | $ax^{a-1}$                                        | $\frac{1}{a+1}x^{a+1}+C$                                     | $a\neq{-1}$                         |
    		  | $a^x$                         | $a^x\ln{a}$                                       | $\frac{1}{\ln{a}}a^x+C$                                      | $a>0, a\neq 1$                      |
    		  | $e^x$                         | $e^x$                                             | $e^x+C$                                                      |                                     |
    		  | $\log_a{x}$                   | $\frac{1}{x\ln{a}}$                               | $x\log_{a}x-\frac{1}{\ln{a}}x+C$                             | 积分使用分部积分法                  |
    		  | $\ln{x}$                      | $\frac{1}{x}$                                     | $x\ln{x}-x+C$                                                | $(\ln|x|)'=\frac{1}{x}$             |
    		  | $\sin{x}$                     | $\cos{x}$                                         | $-\cos{x}+C$                                                 |                                     |
    		  | $\cos{x}$                     | $-\sin{x}$                                        | $\sin{x}+C$                                                  |                                     |
    		  | $\sec{x}$                     | $\sec{x}\tan{x}$                                  | $\ln{|\sec{x}+\tan{x}|}+C$                                   |                                     |
    		  | $\csc{x}$                     | $-\csc{x}\cot{x}$                                 | $-\ln|\csc{x}+\cot{x}|+C=\ln|\csc{x}-\cot{x}|+C$             | NoteA                               |
    		  | $\tan{x}$                     | $\sec^2{x}$                                       | $-\ln|\cos{x}|+C=\ln|\sec{x}|+C$                             |                                     |
    		  | $\cot{x}$                     | $-\csc^2{x}$                                      | $\ln|\sin{x}|+C=-\ln|\csc{x}|+C$                             |                                     |
    		  | $\sin^2{x}$                   |                                                   | $\frac{x}{2}-\frac{\sin{2x}}{4}+C$                           | $\sin^2{x}=\frac{1}{2}(1-\cos{2x})$ |
    		  | $\cos^2{x}$                   |                                                   | $\frac{x}{2}+\frac{\sin{2x}}{4}+C$                           | $\cos^2{x}=\frac{1}{2}(1+\cos{2x})$ |
    		  | $\sec^2{x}$                   |                                                   | $\tan{x}+C$                                                  |                                     |
    		  | $\csc^2{x}$                   |                                                   | $-\cot{x}+C$                                                 |                                     |
    		  | $\tan^2{x}$                   |                                                   | $\tan{x}-x+C$                                                | $\tan^2{x}=\sec^2x-1$               |
    		  | $\cot^2{x}$                   |                                                   | $-\cot{x}-x+C$                                               | $\cot^2{x}=\csc^2{x}-1$             |
    		  | $\sec{x}\tan{x}$              |                                                   | $\sec{x}+C$                                                  |                                     |
    		  | $\csc{x}\cot{x}$              |                                                   | $-\csc{x}+C$                                                 |                                     |
    		  | $\arcsin{x}$                  | $\frac{1}{\sqrt{1-x^2}}$                          |                                                              |                                     |
    		  | $\arccos{x}$                  | $-\frac{1}{\sqrt{1-x^2}}$                         |                                                              |                                     |
    		  | $\arctan{x}$                  | $\frac{1}{1+x^2}$                                 |                                                              |                                     |
    		  | $\mathrm{arc}\cot{x}$         | $-\frac{1}{1+x^2}$                                |                                                              |                                     |
    		  | $\frac{1}{\sqrt{x^2\pm a^2}}$ |                                                   | $\ln|{x}+\sqrt{x^2\pm a^2}|+C$                               |                                     |
    		  | $\frac{1}{\sqrt{a^2-x^2}}$    |                                                   | $\arcsin{\frac{x}{a}}+C$                                     |                                     |
    		  | $\frac{1}{\sqrt{1-x^2}}$      |                                                   | $\arcsin{x}+C$                                               |                                     |
    		  | $\frac{1}{x^2+a^2}$           |                                                   | $\frac{1}{a}\arctan{\frac{x}{a}}+C$                          |                                     |
    		  | $\frac{1}{1+x^2}$             |                                                   | $\arctan{x}+C$                                               |                                     |
    		  | $\frac{1}{x^2-a^2}$           |                                                   | $\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C$                         |                                     |
    		  | $\sqrt{a^2-x^2}$              |                                                   | $\frac{1}{2}(a^2\arcsin{\frac{x}{a}}+x\sqrt{a^2-x^2})+C$     | $(a>|x|\geqslant{0})$               |
    		  | $\sqrt{x^2\pm{}a^2}$          |                                                   | $\frac{1}{2}(x\sqrt{x^2\pm{}a^2}\pm{}a^2\ln|x+\sqrt{x^2\pm{}a^2}|)$ | 参考数学分析(张)                    |
posted @   xuchaoxin1375  阅读(20)  评论(0)  编辑  收藏  举报  
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