I have to say I’m not that type of person who keeps a determined mind on a certain thing, after a month of attempt. If any words appropriate to describe my study, it should be “Always interested in everything, never about to start up with anything.” That’s a shame, indeed.
Ah, another month of rise and fall...
A mindmap for the present study on calculate:
CONCEPTS
FORMAT
INTRODUCTION
limit lim x → x 0 f ( x ) = A \lim\limits_{x\to x_0}f(x)=A x → x 0 lim f ( x ) = A (ϵ − δ ) \epsilon-\delta) ϵ − δ ) ∀ ϵ > 0 , ∃ δ > 0 ; \forall\epsilon > 0, \exist\delta > 0; ∀ ϵ > 0 , ∃ δ > 0 ; 0 < ∣ f ( x ) − A ∣ < ϵ : x ∈ U ˚ ( x 0 , δ ) 0<\vert f(x) - A \vert <\epsilon: x \in \mathring{U}(x_0,\delta) 0 < ∣ f ( x ) − A ∣ < ϵ : x ∈ U ˚ ( x 0 , δ )
right limit f + ( x ) = f ( x + ) = lim x → x 0 + f ( x ) = A f_+(x)=f(x^+)=\lim\limits_{x\to x_0^+}f(x)=A f + ( x ) = f ( x + ) = x → x 0 + lim f ( x ) = A (ϵ − δ ) \epsilon-\delta) ϵ − δ ) ∀ ϵ > 0 , ∃ δ > 0 ; \forall\epsilon > 0, \exist\delta > 0; ∀ ϵ > 0 , ∃ δ > 0 ; 0 < ∣ f ( x ) − A ∣ < ϵ : 0 < x − x 0 < δ 0<\vert f(x) - A \vert <\epsilon: 0<x-x_0<\delta 0 < ∣ f ( x ) − A ∣ < ϵ : 0 < x − x 0 < δ
left limit f − ( x ) = f ( x − ) = lim x → x 0 − f ( x ) = A f_-(x)=f(x^-)=\lim\limits_{x\to x_0^-}f(x)=A f − ( x ) = f ( x − ) = x → x 0 − lim f ( x ) = A (ϵ − δ ) \epsilon-\delta) ϵ − δ ) ∀ ϵ > 0 , ∃ δ > 0 ; \forall\epsilon > 0, \exist\delta > 0; ∀ ϵ > 0 , ∃ δ > 0 ; 0 < ∣ f ( x ) − A ∣ < ϵ : 0 < x 0 − x < δ 0<\vert f(x) - A \vert <\epsilon: 0<x_0-x<\delta 0 < ∣ f ( x ) − A ∣ < ϵ : 0 < x 0 − x < δ
bounded ∣ f ( x ) ∣ < M : \vert f(x)\vert<M: ∣ f ( x ) ∣ < M : M > 0 {M>0} M > 0
infinitesimal
infinity
order
the speed of variable
continuity
lim Δ x → 0 Δ y = 0 \lim\limits_{\Delta x\to0}\Delta y= 0 Δ x → 0 lim Δ y = 0 When x 0 ∈ U ( x 0 , δ ) x_0\in U(x_0,\delta) x 0 ∈ U ( x 0 , δ ) f ( x 0 ) f(x_0) f ( x 0 ) lim x → x 0 f ( x ) \lim\limits_{x\to x_0}f(x) x → x 0 lim f ( x ) f ( x 0 ) = lim x → x 0 f ( x ) f(x_0) = \lim\limits_{x\to x_0}f(x) f ( x 0 ) = x → x 0 lim f ( x )
discontinuity point
derivative
one-sided derivative
inexplicit function
(accurate)derivative
A Δ x + ο ( Δ x ) A\Delta x+\omicron(\Delta x) A Δ x + ο ( Δ x )
(approximate)derivative
f ′ ( x ) d x f'(x)dx f ′ ( x ) d x
approxiamate
f ( x ) ≈ f ′ ( x 0 ) ( x − x 0 ) + f ( x 0 ) f(x)\approx f'(x_0)(x-x_0)+f(x_0) f ( x ) ≈ f ′ ( x 0 ) ( x − x 0 ) + f ( x 0 )
error term
R ( x ) R(x) R ( x )
Peano term
Lagrange term
point of inflection
f ′ ′ ( x ) = 0 : f + ′ ′ ( x ) + f − ′ ′ ( x ) = 0 f''(x)=0:{f''_+(x)+f''_-(x)=0} f ′ ′ ( x ) = 0 : f + ′ ′ ( x ) + f − ′ ′ ( x ) = 0
integrable
f ( x ) = F ′ ( x ) : x ∈ C ( a , b ) f(x)=F'(x):x\in C(a,b) f ( x ) = F ′ ( x ) : x ∈ C ( a , b )
integral upper
Φ ( x ) = ∫ a x f ( x ) d x : \varPhi (x)=\intop^x_af(x)dx: Φ ( x ) = ∫ a x f ( x ) d x : a ⩽ x ⩽ b a\leqslant x\leqslant b a ⩽ x ⩽ b
瑕點
積分元素
積分和
積分變量
積分上限
積分下限
積分區間
被積表達式
夾逼定理 φ ( x ) ⩽ f ( x ) ⩽ ψ ( x ) : x ∈ U ˚ ( x 0 , δ ) , \varphi(x)\leqslant f(x)\leqslant\psi(x):x\in \mathring{U}(x_0,\delta), φ ( x ) ⩽ f ( x ) ⩽ ψ ( x ) : x ∈ U ˚ ( x 0 , δ ) ,
lim x → x 0 φ ( x ) = lim x → x 0 ψ ( x ) = A ; \lim\limits_{x\to x_0}\varphi(x)=\lim\limits_{x\to x_0}\psi(x)=A; x → x 0 lim φ ( x ) = x → x 0 lim ψ ( x ) = A ;
∴ lim x → x 0 f ( x ) = A \therefore \lim\limits_{x\to x_0}f(x)=A ∴ x → x 0 lim f ( x ) = A
同濟大學數學系，高等教育出版社，《高等數學（第七版 上冊）》P46～P47，ISBN 978-7-04-039663-8
零點存在定理 & 介值定理 y = f ( x ) : x ∈ C [ a , b ] , f ( a ) ⋅ f ( b ) < 0 ; y=f(x):x\in \text{C}[a,b],f(a)\cdot f(b)<0; y = f ( x ) : x ∈ C [ a , b ] , f ( a ) ⋅ f ( b ) < 0 ;
∴ f ( ξ ) = 0 : ξ ∈ [ a , b ] \therefore f(\xi)=0:\xi\in[a,b] ∴ f ( ξ ) = 0 : ξ ∈ [ a , b ]
Intermediate Value Theorem介值定理
y = f ( x ) : x ∈ C [ a , b ] , { f ( a ) = A f ( b ) = B ; y=f(x):x\in \text{C}[a,b], \begin{cases} f(a)=A \\f(b)=B \end{cases}; y = f ( x ) : x ∈ C [ a , b ] , { f ( a ) = A f ( b ) = B ;
∴ f ( ξ ) = C : ξ ∈ [ a , b ] , A ⩽ C ⩽ B \therefore f(\xi)=C:\xi\in[a,b],A\leqslant C\leqslant B ∴ f ( ξ ) = C : ξ ∈ [ a , b ] , A ⩽ C ⩽ B
y = f ( x ) : x ∈ C [ a , b ] , { f m a x ( x ) = M f m i n ( x ) = m ; y=f(x):x\in \text{C}[a,b],\begin{cases} f_{max}(x)=M \\f_{min}(x)=m \end{cases}; y = f ( x ) : x ∈ C [ a , b ] , { f m a x ( x ) = M f m i n ( x ) = m ; ∴ R f = [ m , M ] \therefore\text{R}_f=[m,M] ∴ R f = [ m , M ]
同濟大學數學系，高等教育出版社，《高等數學（第七版 上冊）》P68，ISBN 978-7-04-039663-8
萊布尼茲公式 ( u v ) ( n ) = ∑ k = 0 n ( n k ) u n − k v k (uv)^{(n)}=\displaystyle\sum^n_{k=0} \begin{pmatrix}n\\k \end{pmatrix} u^{n-k}v^k ( u v ) ( n ) = k = 0 ∑ n ( n k ) u n − k v k
同濟大學數學系，高等教育出版社，《高等數學（第七版 上冊）》P99，ISBN 978-7-04-039663-8
費馬引理 Diverse functions can be divided according to various standards, yet still some extremely specialized and significant ones which have been studied of sophistication worth command, for their fundamental properties.
There are 6 basic primary functions in total with their derivatives, differentials and integrals ready to employ anywhere and anytime.
constant functionf ( x ) = C : C ∈ R f(x)=C:C\in R f ( x ) = C : C ∈ R
power functionf ( x ) = x n : n ∈ R f(x)=x^n:n\in R f ( x ) = x n : n ∈ R
exponential function
ordinary statef ( x ) = a x : a ∈ R f(x)=a^x:a\in R f ( x ) = a x : a ∈ R
natural constant statef ( x ) = e x f(x)=e^x f ( x ) = e x
logarithmic function
ordinary statef ( x ) = log a x : a ∈ R f(x)=\log_a x:a\in R f ( x ) = log a x : a ∈ R
natural constant statef ( x ) = ln x f(x)=\ln x f ( x ) = ln x
decuple statef ( x ) = lg x f(x)=\lg x f ( x ) = lg x
trigonometric functions
sine f ( x ) = sin x f(x)=\sin x f ( x ) = sin x
tangent f ( x ) = tan x f(x)=\tan x f ( x ) = tan x
secant f ( x ) = sec x f(x)=\sec x f ( x ) = sec x
cosine f ( x ) = cos x f(x)=\cos x f ( x ) = cos x
cotangent f ( x ) = cot x f(x)=\cot x f ( x ) = cot x
cosecant f ( x ) = csc x f(x)=\csc x f ( x ) = csc x
inverse trigonometric functions
arcsine f ( x ) = arcsin x f(x)=\arcsin x f ( x ) = arcsin x
arc cosine f ( x ) = arccos x f(x)=\arccos x f ( x ) = arccos x
arctangent f ( x ) = arctan x f(x)=\arctan x f ( x ) = arctan x
hyperbolic sinef ( x ) = sh x = e x − e − x 2 f(x)=\sh x=\frac{\text{e}^x-\text{e}^{-x}}{2} f ( x ) = sh x = 2 e x − e − x
hyperbolic cosinef ( x ) = ch x = e x + e − x 2 f(x)=\ch x=\frac{\text{e}^x+\text{e}^{-x}}{2} f ( x ) = ch x = 2 e x + e − x
hyperbolic tangentf ( x ) = th x = e x − e − x e x + e − x f(x)=\th x=\frac{\text{e}^x-\text{e}^{-x}}{\text{e}^x+\text{e}^{-x}} f ( x ) = th x = e x + e − x e x − e − x
inverse hyperbolic sinef ( x ) = arsh x = ln ( x + x 2 + 1 ) f(x)=\text{arsh}\ x=\ln\Big(x+\sqrt{x^2+1}\Big) f ( x ) = arsh x = ln ( x + x 2 + 1 )
inverse hyperbolic cosinef ( x ) = arch x = ln ( x + x 2 − 1 ) f(x)=\text{arch}\ x=\ln\Big(x+\sqrt{x^2-1}\Big) f ( x ) = arch x = ln ( x + x 2 − 1 )
inverse hyperbolic tangentf ( x ) = arth x = 1 2 ln 1 + x 1 − x f(x)=\text{arth}\ x=\frac{1}{2}\ln\frac{1+x}{1-x} f ( x ) = arth x = 2 1 ln 1 − x 1 + x
x ∼ sin x ∼ arcsin x ∼ tan x ∼ arctan x ∼ ln ( 1 + x ) ∼ e x − 1 x \sim \sin x \sim \arcsin x\sim \tan x\sim \arctan x\sim\ln(1+x)\sim\text{e}^x-1 x ∼ sin x ∼ arcsin x ∼ tan x ∼ arctan x ∼ ln ( 1 + x ) ∼ e x − 1 1 − cos x ∼ x 2 2 1-\cos x\sim\frac{x^2}{2} 1 − cos x ∼ 2 x 2 1 + x n − 1 ∼ x n \sqrt[n]{1+x}-1\sim\frac{x}{n} n 1 + x − 1 ∼ n x a x − 1 ∼ x ln a a^x-1\sim x\ln a a x − 1 ∼ x ln a ( 1 + x ) m − 1 ∼ m x (1+x)^m-1\sim mx ( 1 + x ) m − 1 ∼ m x tan x − sin x ∼ x 3 2 \tan x-\sin x\sim\frac{x^3}{2} tan x − sin x ∼ 2 x 3
lim x → 0 sin x x = 1 \lim\limits_{x\to0}\frac{\sin x}{x}=1 x → 0 lim x sin x = 1 lim x → ∞ ( 1 + 1 x ) x = e \lim\limits_{x\to\infin}(1+\frac{1}{x})^x=e x → ∞ lim ( 1 + x 1 ) x = e Llimit and Rlimit exist
else
Llimit equivalent to Rlimit
else
limit equivalent to Infinite
infinitely oscillating limit
Discontinuity Point
The First Class
The Seconde Class
Removable Discontinuity Point
Jump Discontinuity Point
Infinite Discontinuity Point
Oscillating Discontinuity Point
TYPE
DEFINATION
LEFT LIMIT & RIGHT LIMIT
Jump Defined
f + ( x ) ≠ f − ( x ) f_+(x)\neq f_-(x) f + ( x ) ̸ = f − ( x )
Removable Defined
f + ( x ) = f − ( x ) f_+(x)=f_-(x) f + ( x ) = f − ( x )
Infinite NOT Defined
lim x → x 0 f ( x ) = ∞ \lim\limits_{x\to x_0} f(x)=\infty x → x 0 lim f ( x ) = ∞
Oscillating NOT Defined
~
Discontinuity Point of the First Class
Defined
Discontinuity Point of the Second Class
NOT Defined
Warning: differential employed for its transformation with ease to derivative as the following.d ( f ( x ) ) d x = f ′ ( x ) ⋅ d x d x = f ′ ( x ) \frac{\text{d}\big(f(x)\big)}{\text{d}x}=\frac{f'(x)\cdot \text{d}x}{\text{d}x}=f'(x) d x d ( f ( x ) ) = d x f ′ ( x ) ⋅ d x = f ′ ( x )
constant differentiald ( C ) = 0 \text{d}(C)=0 d ( C ) = 0
power differentiald ( x a ) = a ⋅ x a − 1 d x \text{d}(x^a)=a\cdot x^{a-1}\text{d}x d ( x a ) = a ⋅ x a − 1 d x
exponential differentiald ( a x ) = a x ⋅ ln a d x \text{d}(a^x)=a^x\cdot \ln a\text{d}x d ( a x ) = a x ⋅ ln a d x d ( e x ) = e x d x \text{d}(\text{e}^x)=\text{e}^x\text{d}x d ( e x ) = e x d x
logarithm differentiald ( log a x ) = d x x ⋅ ln a \text{d}(\log _ax)=\frac{\text{d}x}{x\cdot \ln a} d ( log a x ) = x ⋅ ln a d x d ( ln x ) = 1 x \text{d}(\ln x)=\frac{1}{x} d ( ln x ) = x 1
trigonometric differentialsd ( sin x ) = cos x d x \text{d}(\sin x)=\cos x\text{d}x d ( sin x ) = cos x d x d ( cos x ) = − sin x d x \text{d}(\cos x)=-\sin x\text{d}x d ( cos x ) = − sin x d x d ( tan x ) = sec 2 x d x \text{d}(\tan x)=\sec ^2x\text{d}x d ( tan x ) = sec 2 x d x d ( cot x ) = − csc 2 x d x \text{d}(\cot x)=-\csc ^2x\text{d}x d ( cot x ) = − csc 2 x d x d ( csc x ) = cot x ⋅ csc x d x \text{d}(\csc x)=\cot x\cdot\csc x\text{d}x d ( csc x ) = cot x ⋅ csc x d x d ( sec x ) = − tan x ⋅ sec x d x \text{d}(\sec x)=-\tan x\cdot\sec x\text{d}x d ( sec x ) = − tan x ⋅ sec x d x
inverse trigonometric differentialsd ( arcsin x ) = d x 1 − x 2 \text{d}(\arcsin x)=\frac{\text{d}x}{\sqrt{1-x^2}} d ( arcsin x ) = 1 − x 2 d x d ( arccos x ) = − d x 1 − x 2 \text{d}(\arccos x)=-\frac{\text{d}x}{\sqrt{1-x^2}} d ( arccos x ) = − 1 − x 2 d x d ( arctan x ) = d x 1 + x 2 \text{d}(\arctan x)=\frac{\text{d}x}{1+x^2} d ( arctan x ) = 1 + x 2 d x
additived ( u ± v ) = d u ± d v \text{d}(u\pm v)=\text{d}u\pm \text{d}v d ( u ± v ) = d u ± d v
multiplicatived ( u ⋅ v ) = v d u + u d v \text{d}(u\cdot v)=v\text{d}u+u\text{d}v d ( u ⋅ v ) = v d u + u d v d ( u v ) = d ( u ⋅ 1 v ) = 1 v d u − u v 2 d v = v d u − u d v v 2 \text{d}\bigg(\frac{u}{v}\bigg)=\text{d}(u\cdot \frac{1}{v})=\frac{1}{v}\text{d}u-\frac{u}{v^2}\text{d}v=\frac{v\text{d}u-u\text{d}v}{v^2} d ( v u ) = d ( u ⋅ v 1 ) = v 1 d u − v 2 u d v = v 2 v d u − u d v
scalar multipliable( a ⋅ x ) = a ⋅ d x : a ∈ R \text(a\cdot x)=a\cdot\text{d}x:a\in R ( a ⋅ x ) = a ⋅ d x : a ∈ R
additive( f ( x ) ± g ( x ) ) ′ = f ′ ( x ) ± g ′ ( x ) \big( f(x)\pm g(x)\big)'=f'(x)\pm g'(x) ( f ( x ) ± g ( x ) ) ′ = f ′ ( x ) ± g ′ ( x )
multiplicative( f ( x ) ⋅ g ( x ) ) = f ′ ( x ) ⋅ g ( x ) + f ( x ) ⋅ g ′ ( x ) \big( f(x)\cdotp g(x)\big)=f'(x)\cdotp g(x)+f(x)\cdotp g'(x) ( f ( x ) ⋅ g ( x ) ) = f ′ ( x ) ⋅ g ( x ) + f ( x ) ⋅ g ′ ( x ) ( f ( x ) g ( x ) ) ′ = ( f ( x ) ⋅ 1 g ( x ) ) ′ = f ′ ( x ) ⋅ 1 g ( x ) − f ( x ) ⋅ g ′ ( x ) g 2 ( x ) = f ′ ( x ) ⋅ g ( x ) − f ( x ) ⋅ g ′ ( x ) g 2 ( x ) \Bigg(\frac{f(x)}{g(x)}\Bigg)'=\big( f(x)\cdotp \frac{1}{g(x)}\big)'=f'(x)\cdotp \frac{1}{g(x)}-f(x)\cdotp \frac{g'(x)}{g^2(x)}=\frac{f'(x)\cdotp g(x)-f(x)\cdotp g'(x)}{g^2(x)} ( g ( x ) f ( x ) ) ′ = ( f ( x ) ⋅ g ( x ) 1 ) ′ = f ′ ( x ) ⋅ g ( x ) 1 − f ( x ) ⋅ g 2 ( x ) g ′ ( x ) = g 2 ( x ) f ′ ( x ) ⋅ g ( x ) − f ( x ) ⋅ g ′ ( x )
scalar multiplication( a ⋅ f ( x ) ) ′ = a ⋅ f ′ ( x ) : a ∈ R \big(a\cdot f(x)\big)'=a\cdot f'(x):a\in R ( a ⋅ f ( x ) ) ′ = a ⋅ f ′ ( x ) : a ∈ R
Warning: derivative employed for appropriate simplification, and it could be transformed with ease to differential as the following.
d ( f ( x ) ) = f ′ ( x ) ⋅ d x \text{d}\big(f(x)\big)=f'(x)\cdot \text{d}x d ( f ( x ) ) = f ′ ( x ) ⋅ d x
( x e x ) ′ = ( x + 1 ) e x (x\text{e}^x)'=(x+1)\text{e}^x ( x e x ) ′ = ( x + 1 ) e x ( x ln x ) ′ = 1 + ln x (x\ln x)'=1+\ln x ( x ln x ) ′ = 1 + ln x
( 1 + x ) a ≈ 1 + a x (1+x)^a \approx 1+ax ( 1 + x ) a ≈ 1 + a x sin x ≈ x \sin x \approx x sin x ≈ x x x x tan x ≈ x \tan x \approx x tan x ≈ x x x x e x ≈ 1 + x \text{e}^x\approx 1+x e x ≈ 1 + x ln ( 1 + x ) ≈ x \ln (1+x)\approx x ln ( 1 + x ) ≈ x
Roller Mean Value Theorem
Lagrange Mean Value Theorem
Cauchy Mean Value Theorem
Taylor Mean Value Theorem
∫ f ( x ) d x = F ( x ) + C \int f(x)\text{d}x=F(x)+C ∫ f ( x ) d x = F ( x ) + C
The equation above matters under the following conditions:
F ′ ( x ) = f ( x ) F'(x)=f(x) F ′ ( x ) = f ( x )
which is to say,
∫ a b f ( x ) d x = ∑ i = 1 n S i = ∑ i = 1 n f ( ξ i ) Δ x i \int _a^bf(x)\text{d}x=\sum_{i=1}^{n}S_i=\sum_{i=1}^{n}f(\xi_i)\Delta x_i ∫ a b f ( x ) d x = i = 1 ∑ n S i = i = 1 ∑ n f ( ξ i ) Δ x i
sin 2 x + cos 2 x = 1 \sin^2 x+\cos^2 x=1 sin 2 x + cos 2 x = 1 tan 2 x + 1 = sec 2 x \tan^2 x+1=\sec^2 x tan 2 x + 1 = sec 2 x ( tan x ) ′ = sec 2 x (\tan x)'=\sec^2 x ( tan x ) ′ = sec 2 x
Warning: The following integral formulas are divided according to the left value.
constant function∫ 0 d x = C \int 0\text{d}x=C ∫ 0 d x = C
power function∫ x a d x = x a + 1 a + 1 + C \int x^a\text{d}x=\frac{x^{a+1}}{a+1}+C ∫ x a d x = a + 1 x a + 1 + C
exponential function∫ a x d x = a x ln a + C \int {a^x} \text{d}x=\frac{a^x}{\ln a}+C ∫ a x d x = ln a a x + C ∫ e x d x = e x + C \int {\text{e}^x} \text{d}x=\text{e}^x+C ∫ e x d x = e x + C
logarithm function (natural constant state)∫ 1 x d x = ln ∣ x ∣ + C \int \frac{1}{x}\text{d}x=\ln \vert x\vert+C ∫ x 1 d x = ln ∣ x ∣ + C
trigonometric functions∫ cos x d x = sin x + C \int \cos x\text{d}x=\sin x+C ∫ cos x d x = sin x + C ∫ sin x d x = − cos x + C \int \sin x\text{d}x=-\cos x +C ∫ sin x d x = − cos x + C ∫ sec 2 x d x = tan x + C \int \sec ^2x\text{d}x=\tan x+C ∫ sec 2 x d x = tan x + C ∫ csc 2 x d x = − cot x + C \int \csc ^2x\text{d}x=-\cot x+C ∫ csc 2 x d x = − cot x + C ∫ csc x cot x d x = csc x + C \int \csc x\cot x\text{d}x=\csc x+C ∫ csc x cot x d x = csc x + C ∫ sec x tan x d x = − sec x + C \int \sec x\tan x\text{d}x=-\sec x+C ∫ sec x tan x d x = − sec x + C
inverse trigonometric functions∫ d x 1 − x 2 = arcsin x + C = − arccos x + C \int \frac{\text{d}x}{\sqrt{1-x^2}}=\arcsin x+C=-\arccos x+C ∫ 1 − x 2 d x = arcsin x + C = − arccos x + C ∫ d x a 2 − x 2 = ∫ d ( x a ) 1 − ( x a ) 2 = arcsin x a + C = − arccos x a + C \int \frac{\text{d}x}{\sqrt{a^2-x^2}}=\int \frac{\text{d}\big(\frac{x}{a}\big)}{\sqrt{1-\big(\frac{x}{a}\big)^2}}=\arcsin\frac{x}{a}+C=-\arccos\frac{x}{a}+C ∫ a 2 − x 2 d x = ∫ 1 − ( a x ) 2 d ( a x ) = arcsin a x + C = − arccos a x + C ∫ d x 1 + x 2 = arctan x + C \int \frac{\text{d}x}{1+x^2}=\arctan x+C ∫ 1 + x 2 d x = arctan x + C ∫ d x a 2 + x 2 = 1 a ∫ d ( x a ) 1 + ( x a ) 2 = 1 a arctan x + C \int \frac{\text{d}x}{a^2+x^2}=\frac{1}{a}\int \frac{{\text{d}\big(\frac{x}{a}\big)}}{1+\big(\frac{x}{a}\big)^2}=\frac{1}{a}\arctan x+C ∫ a 2 + x 2 d x = a 1 ∫ 1 + ( a x ) 2 d ( a x ) = a 1 arctan x + C
∫ f ( x ) d x → u = φ ( x ) ∫ f ( φ − 1 ( u ) ) ⋅ φ ′ ( x ) d x φ ′ ( x ) = 1 φ ′ ( x ) ∫ g ( u ) d u \int f(x)\text{d}x\xrightarrow{u=\varphi(x)}\int f(\varphi ^{-1}(u))\cdot\frac{\varphi'(x)\text{d}x}{\varphi'(x)}=\frac{1}{\varphi '(x)}\int g(u)\text{d}u ∫ f ( x ) d x u = φ ( x ) ∫ f ( φ − 1 ( u ) ) ⋅ φ ′ ( x ) φ ′ ( x ) d x = φ ′ ( x ) 1 ∫ g ( u ) d u
∫ a b f ( x ) d x → α = φ ( a ) , β = φ ( b ) u = φ ( x ) ( x = φ − 1 ( u ) ) ∫ α β f ( φ − 1 ( u ) ) ⋅ φ ′ ( x ) d x φ ′ ( x ) = 1 φ ′ ( x ) ∫ α β g ( u ) d u \int_a^b f(x)\text{d}x\xrightarrow[\alpha=\varphi(a),\beta=\varphi(b)]{u=\varphi(x)\big(x=\varphi^{-1}(u)\big)}\int_{\alpha}^{\beta} f(\varphi ^{-1}(u))\cdot\frac{\varphi'(x)\text{d}x}{\varphi'(x)}=\frac{1}{\varphi '(x)}\int_{\alpha}^{\beta} g(u)\text{d}u ∫ a b f ( x ) d x u = φ ( x ) ( x = φ − 1 ( u ) ) α = φ ( a ) , β = φ ( b ) ∫ α β f ( φ − 1 ( u ) ) ⋅ φ ′ ( x ) φ ′ ( x ) d x = φ ′ ( x ) 1 ∫ α β g ( u ) d u
∫ f ( x ) d x → x = ψ ( t ) ∫ f ( ψ ( t ) ) d ( ψ ( t ) ) = ψ ′ ( t ) ∫ g ( t ) d t \int f(x)\text{d}x\xrightarrow{x=\psi(t)}\int f\big(\psi(t)\big)\text{d}\big(\psi(t)\big)=\psi'(t)\int g(t)\text{d}t ∫ f ( x ) d x x = ψ ( t ) ∫ f ( ψ ( t ) ) d ( ψ ( t ) ) = ψ ′ ( t ) ∫ g ( t ) d t
∫ κ ζ f ( x ) d x → κ = ψ ( k ) , ζ = ψ ( l ) ( k = ψ − 1 ( κ ) , l = ψ − 1 ( ζ ) ) x = ψ ( t ) ∫ k l f ( ψ ( t ) ) d ( ψ ( t ) ) = ψ ′ ( t ) ∫ k l g ( t ) d t \int_{\kappa}^{\zeta}
f(x)\text{d}x\xrightarrow[\kappa=\psi(k),\zeta=\psi(l) \big(k=\psi^{-1}(\kappa),l=\psi^{-1}(\zeta)\big)]{x=\psi(t)}\int_{k}^{l} f\big(\psi(t)\big)\text{d}\big(\psi(t)\big)=\psi'(t)\int_{k}^{l} g(t)\text{d}t ∫ κ ζ f ( x ) d x x = ψ ( t ) κ = ψ ( k ) , ζ = ψ ( l ) ( k = ψ − 1 ( κ ) , l = ψ − 1 ( ζ ) ) ∫ k l f ( ψ ( t ) ) d ( ψ ( t ) ) = ψ ′ ( t ) ∫ k l g ( t ) d t
∫ u d v = u v + ∫ v d u \int u\text{d}v=uv+\int v\text{d}u ∫ u d v = u v + ∫ v d u
∫ x d x = ∫ x 1 2 d x = 2 3 ∫ 3 2 x 1 2 d x = 2 3 ∫ ( x 3 2 ) ′ d x = 2 3 x 3 2 d x + C \int\sqrt x\text{d}x=\int x^{\frac{1}{2}}\text{d}x=\frac{2}{3}\int \frac{3}{2}x^{\frac{1}{2}}\text{d}x=\frac{2}{3}\int (x^{\frac{3}{2}})'\text{d}x=\frac{2}{3}x^{\frac{3}{2}}\text{d} x+C ∫ x d x = ∫ x 2 1 d x = 3 2 ∫ 2 3 x 2 1 d x = 3 2 ∫ ( x 2 3 ) ′ d x = 3 2 x 2 3 d x + C
∫ d x x = ∫ x − 1 2 d x = 2 ∫ 1 2 x − 1 2 = 2 ∫ ( x 1 2 ) ′ d x = 2 x 1 2 d x + C \int\frac{\text{d} x}{\sqrt x}=\int x^{-\frac{1}{2}}\text{d}x=2\int \frac{1}{2}x^{-\frac{1}{2}}=2\int(x^{\frac{1}{2}})'\text{d}x=2x^{\frac{1}{2}}\text{d}x+C ∫ x d x = ∫ x − 2 1 d x = 2 ∫ 2 1 x − 2 1 = 2 ∫ ( x 2 1 ) ′ d x = 2 x 2 1 d x + C
∫ tan x d x = ∫ sin x cos x d x = ∫ − d ( cos x ) cos x = − ln ∣ cos x ∣ + C \int \tan x\text{d}x=\int \frac{\sin x}{\cos x}\text{d}x=\int -\frac{\text{d}(\cos x)}{\cos x}=-\ln\vert\cos x\vert+C ∫ tan x d x = ∫ cos x sin x d x = ∫ − cos x d ( cos x ) = − ln ∣ cos x ∣ + C
∫ cot x d x = ∫ cos x sin x d x = ∫ d ( sin x ) sin x = ln ∣ sin x ∣ + C \int \cot x\text{d}x=\int \frac{\cos x}{\sin x}\text{d}x=\int \frac{\text{d}(\sin x)}{\sin x}=\ln \vert\sin x\vert+C ∫ cot x d x = ∫ sin x cos x d x = ∫ sin x d ( sin x ) = ln ∣ sin x ∣ + C
∫ d x 1 − x 2 = ∫ 1 2 ( 1 1 − x + 1 1 + x ) d x = 1 2 ( ∫ − d ( 1 − x ) 1 − x + ∫ d ( 1 + x ) 1 + x ) = 1 2 ( − ln ( 1 − x ) + ln ( 1 + x ) ) + C = 1 2 ln ∣ 1 + x 1 − x ∣ + C \int \frac{\text{d}x}{1-x^2}=\int\frac{1}{2}\Bigg(\frac{1}{1-x}+\frac{1}{1+x}\Bigg)\text{d}x=\frac{1}{2}\Bigg(\int-\frac{\text{d}(1-x)}{1-x}+\int\frac{\text{d}(1+x)}{1+x}\Bigg)=\frac{1}{2}\Big(-\ln(1-x)+\ln(1+x)\Big)+C=\frac{1}{2}\ln\vert\frac{1+x}{1-x}\vert+C ∫ 1 − x 2 d x = ∫ 2 1 ( 1 − x 1 + 1 + x 1 ) d x = 2 1 ( ∫ − 1 − x d ( 1 − x ) + ∫ 1 + x d ( 1 + x ) ) = 2 1 ( − ln ( 1 − x ) + ln ( 1 + x ) ) + C = 2 1 ln ∣ 1 − x 1 + x ∣ + C
∫ d x a 2 − x 2 = 1 a ∫ d ( x a ) 1 − ( x a ) 2 = 1 a ∫ 1 2 ( 1 1 − x a + 1 1 + x a ) d ( x a ) = 1 2 a ( ∫ d ( x a ) 1 − x a + ∫ d ( x a ) 1 + x a ) = 1 2 a ( ∫ − d ( 1 − x a ) 1 − x a + ∫ d ( 1 + x a ) 1 + x a ) = 1 2 a ( − ln ∣ 1 − x a ∣ + ln ∣ 1 + x a ∣ ) + C = 1 2 a ln ∣ 1 + x a 1 − x a ∣ + C = 1 2 a ln ∣ a + x a − x ∣ + C \int \frac{\text{d}x}{a^2-x^2}=\frac{1}{a}\int\frac{\text{d}\big(\frac{x}{a}\big)}{1-\big(\frac{x}{a}\big)^2}=\frac{1}{a}\int\frac{1}{2}\Bigg(\frac{1}{1-\frac{x}{a}}+\frac{1}{1+\frac{x}{a}}\Bigg)\text{d}\Bigg(\frac{x}{a}\Bigg)=\frac{1}{2a}\Bigg(\int\frac{\text{d}\big(\frac{x}{a}\big)}{1-\frac{x}{a}}+\int\frac{\text{d}\big(\frac{x}{a}\big)}{1+\frac{x}{a}}\Bigg)=\frac{1}{2a}\Bigg(\int-\frac{\text{d}\big(1-\frac{x}{a}\big)}{1-\frac{x}{a}}+\int\frac{\text{d}\big(1+\frac{x}{a}\big)}{1+\frac{x}{a}}\Bigg)=\frac{1}{2a}\Bigg(-\ln\vert 1-\frac{x}{a}\vert+\ln\vert 1+\frac{x}{a}\vert\Bigg)+C=\frac{1}{2a}\ln\vert\frac{1+\frac{x}{a}}{1-\frac{x}{a}}\vert+C=\frac{1}{2a}\ln\vert\frac{a+x}{a-x}\vert+C ∫ a 2 − x 2 d x = a 1 ∫ 1 − ( a x ) 2 d ( a x ) = a 1 ∫ 2 1 ( 1 − a x 1 + 1 + a x 1 ) d ( a x ) = 2 a 1 ( ∫ 1 − a x d ( a x ) + ∫ 1 + a x d ( a x ) ) = 2 a 1 ( ∫ − 1 − a x d ( 1 − a x ) + ∫ 1 + a x d ( 1 + a x ) ) = 2 a 1 ( − ln ∣ 1 − a x ∣ + ln ∣ 1 + a x ∣ ) + C = 2 a 1 ln ∣ 1 − a x 1 + a x ∣ + C = 2 a 1 ln ∣ a − x a + x ∣ + C
∫ sec x d x = ∫ ln ∣ sec x + tan x ∣ + C \int \sec x\text{d}x=\int \ln\vert\sec x+\tan x\vert +C ∫ sec x d x = ∫ ln ∣ sec x + tan x ∣ + C
∫ csc x d x = ∫ ln ∣ csc x − cot x ∣ + C \int \csc x\text{d}x=\int \ln\vert\csc x-\cot x\vert +C ∫ csc x d x = ∫ ln ∣ csc x − cot x ∣ + C
∫ d x x 2 ± a 2 = 1 a ln ∣ x + x 2 + a 2 ∣ + C \int \frac{\text{d}x}{\sqrt{x^2\pm a^2}}=\frac{1}{a}\ln\vert x+\sqrt{x^2+a^2}\vert +C ∫ x 2 ± a 2 d x = a 1 ln ∣ x + x 2 + a 2 ∣ + C
∫ x e x d x = ( x − 1 ) e x + C \int x\text{e}^x\text{d}x=(x-1)\text{e}^x+C ∫ x e x d x = ( x − 1 ) e x + C
∫ x e − x d x = − ( x + 1 ) e x + C \int x\text{e}^{-x}\text{d}x=-(x+1)\text{e}^x+C ∫ x e − x d x = − ( x + 1 ) e x + C
∫ sin x e x d x = 1 2 ( sin x − cos x ) e x + C \int \sin x\text{e}^x\text{d}x=\frac{1}{2}({\sin x-\cos x})\text{e}^x+C ∫ sin x e x d x = 2 1 ( sin x − cos x ) e x + C
I n = x n e x − I n − 1 : I n = ∫ x n e x d x , I 1 = ∫ x e x d x = ( x − 1 ) e x + C I_n=x^n\text{e}^x-I_{n-1}:I_n=\int x^n\text{e}^x\text{d}x,I_1=\int x\text{e}^x\text{d}x=(x-1)\text{e}^x+C I n = x n e x − I n − 1 : I n = ∫ x n e x d x , I 1 = ∫ x e x d x = ( x − 1 ) e x + C
∫ ln x d x = x ln x − x + C \int \ln x\text{d}x=x\ln x-x+C ∫ ln x d x = x ln x − x + C
