Values ​​of antiderivatives table. Primitive. Solving easy examples

Definition of antiderivative function

• Function y=F(x) is called the antiderivative for the function y=f(x) at a given interval X, if for all X ∈X equality holds: F′(x) = f(x)

It can be read in two ways:

1. f function derivative F
2. F antiderivative for function f

property of antiderivatives

• If F(x)- antiderivative for the function f(x) on a given interval, then the function f(x) has infinitely many antiderivatives, and all these antiderivatives can be written as F(x) + C, where C is an arbitrary constant.

Geometric interpretation

• Graphs of all antiderivatives of a given function f(x) are obtained from the graph of any one antiderivative by parallel translations along the O axis at.

Rules for computing antiderivatives

1. The antiderivative of the sum is equal to the sum of the antiderivatives. If F(x)- primitive for f(x), and G(x) is the antiderivative for g(x), That F(x) + G(x)- primitive for f(x) + g(x).
2. The constant factor can be taken out of the sign of the derivative. If F(x)- primitive for f(x), And k is constant, then kF(x)- primitive for kf(x).
3. If F(x)- primitive for f(x), And k,b- permanent, and k ≠ 0, That 1/k F(kx + b)- primitive for f(kx + b).

Remember!

Any function F (x) \u003d x 2 + C , where C is an arbitrary constant, and only such a function is an antiderivative for the function f(x) = 2x.

• For example:

  F "(x) \u003d (x 2 + 1)" \u003d 2x \u003d f (x);

  f(x) = 2x, because F "(x) \u003d (x 2 - 1)" \u003d 2x \u003d f (x);

  f(x) = 2x, because F "(x) \u003d (x 2 -3)" \u003d 2x \u003d f (x);

Relationship between graphs of a function and its antiderivative:

1. If the graph of the function f(x)>0 on the interval, then the graph of its antiderivative F(x) increases over this interval.
2. If the graph of the function f(x) on the interval, then the graph of its antiderivative F(x) decreases over this interval.
3. If f(x)=0, then the graph of its antiderivative F(x) at this point changes from increasing to decreasing (or vice versa).

To denote the antiderivative, the sign of the indefinite integral is used, that is, the integral without indicating the limits of integration.

Indefinite integral

Definition:

• The indefinite integral of the function f(x) is the expression F(x) + C, that is, the set of all antiderivatives of the given function f(x). The indefinite integral is denoted as follows: \int f(x) dx = F(x) + C

• f(x) is called the integrand;
• f(x) dx- is called the integrand;
• x- is called the variable of integration;
• F(x)- one of the antiderivatives of the function f(x);
• WITH is an arbitrary constant.

Properties of the indefinite integral

1. The derivative of the indefinite integral is equal to the integrand: (\int f(x) dx)\prime= f(x) .
2. The constant factor of the integrand can be taken out of the integral sign: \int k \cdot f(x) dx = k \cdot \int f(x) dx.
3. The integral of the sum (difference) of functions is equal to the sum (difference) of the integrals of these functions: \int (f(x) \pm g(x)) dx = \int f(x) dx \pm \int g(x) dx.
4. If k,b are constants, and k ≠ 0, then \int f(kx + b) dx = \frac ( 1 ) ( k ) \cdot F(kx + b) + C.

Table of antiderivatives and indefinite integrals

Function f(x)	antiderivative F(x) + C	Indefinite integrals \int f(x) dx = F(x) + C
0	C	\int 0 dx = C
f(x) = k	F(x) = kx + C	\int kdx = kx + C
f(x) = x^m, m\not =-1	F(x) = \frac ( x^ ( m+1 ) ) ( m+1 ) + C	\int x ( ^m ) dx = \frac ( x^ ( m+1 ) ) ( m+1 ) + C
f(x) = \frac ( 1 ) ( x )	F(x) = l n \lvert x \rvert + C	\int \frac ( dx ) ( x ) = l n \lvert x \rvert + C
f(x) = e^x	F(x) = e^x + C	\int e ( ^x ) dx = e^x + C
f(x) = a^x	F(x) = \frac ( a^x ) ( lna ) + C	\int a ( ^x ) dx = \frac ( a^x ) ( l na ) + C
f(x) = \sin x	F(x) = -\cos x + C	\int \sin x dx = -\cos x + C
f(x) = \cos x	F(x)=\sin x + C	\int \cos x dx = \sin x + C
f(x) = \frac ( 1 ) ( \sin ( ^2 ) x )	F(x) = -\ctg x + C	\int \frac ( dx ) ( \sin ( ^2 ) x ) = -\ctg x + C
f(x) = \frac ( 1 ) ( \cos ( ^2 ) x )	F(x) = \tg x + C	\int \frac ( dx ) ( \sin ( ^2 ) x ) = \tg x + C
f(x) = \sqrt ( x )	F(x) =\frac ( 2x \sqrt ( x ) ) ( 3 ) + C
f(x) =\frac ( 1 ) ( \sqrt ( x ) )	F(x) =2\sqrt ( x ) + C
f(x) =\frac ( 1 ) ( \sqrt ( 1-x^2 ) )	F(x)=\arcsin x + C	\int \frac ( dx ) ( \sqrt ( 1-x^2 ) ) =\arcsin x + C
f(x) =\frac ( 1 ) ( \sqrt ( 1+x^2 ) )	F(x)=\arctg x + C	\int \frac ( dx ) ( \sqrt ( 1+x^2 ) ) =\arctg x + C
f(x)=\frac ( 1 ) ( \sqrt ( a^2-x^2 ) )	F(x)=\arcsin \frac ( x ) ( a ) + C	\int \frac ( dx ) ( \sqrt ( a^2-x^2 ) ) =\arcsin \frac ( x ) ( a ) + C
f(x)=\frac ( 1 ) ( \sqrt ( a^2+x^2 ) )	F(x)=\arctg \frac ( x ) ( a ) + C	\int \frac ( dx ) ( \sqrt ( a^2+x^2 ) ) = \frac ( 1 ) ( a ) \arctg \frac ( x ) ( a ) + C
f(x) =\frac ( 1 ) ( 1+x^2 )	F(x)=\arctg + C	\int \frac ( dx ) ( 1+x^2 ) =\arctg + C
f(x)=\frac ( 1 ) ( \sqrt ( x^2-a^2 ) ) (a \not= 0)	F(x)=\frac ( 1 ) ( 2a ) l n \lvert \frac ( x-a ) ( x+a ) \rvert + C	\int \frac ( dx ) ( \sqrt ( x^2-a^2 ) ) =\frac ( 1 ) ( 2a ) l n \lvert \frac ( x-a ) ( x+a ) \rvert + C
f(x)=\tg x	F(x)= - l n \lvert \cos x \rvert + C	\int \tg x dx =-l n \lvert \cos x \rvert + C
f(x)=\ctg x	F(x)= l n \lvert \sin x \rvert + C	\int \ctg x dx = l n \lvert \sin x \rvert + C
f(x)=\frac ( 1 ) ( \sin x )	F(x)= l n \lvert \tg \frac ( x ) ( 2 ) \rvert + C	\int \frac ( dx ) ( \sin x ) = l n \lvert \tg \frac ( x ) ( 2 ) \rvert + C
f(x)=\frac ( 1 ) ( \cos x )	F(x)= l n \lvert \tg (\frac ( x ) ( 2 ) +\frac ( \pi ) ( 4 )) \rvert + C	\int \frac ( dx ) ( \cos x ) = l n \lvert \tg (\frac ( x ) ( 2 ) +\frac ( \pi ) ( 4 )) \rvert + C

Newton–Leibniz formula

Let f(x) this function, F its arbitrary primitive.

\int_ ( a ) ^ ( b ) f(x) dx =F(x)|_ ( a ) ^ ( b )= F(b) - F(a)

Where F(x)- primitive for f(x)

That is, the integral of the function f(x) on the interval is equal to the difference of the antiderivatives at the points b And a.

Area of ​​a curvilinear trapezoid

Curvilinear trapezoid is called a figure limited schedule non-negative and continuous on the segment of the function f, axis Ox and straight lines x = a And x = b.

The area of ​​a curvilinear trapezoid is found using the Newton-Leibniz formula:

S= \int_ ( a ) ^ ( b ) f(x) dx

Direct integration using the table of antiderivatives (tables of indefinite integrals)

Table of antiderivatives

We can find the antiderivative by the known differential of a function if we use the properties of the indefinite integral. From the table of basic elementary functions, using the equalities ∫ d F (x) = ∫ F "(x) d x = ∫ f (x) d x = F (x) + C and ∫ k f (x) d x = k ∫ f (x) d x we ​​can make a table of antiderivatives.

We write the table of derivatives in the form of differentials.

Constant y = C C" = 0 Power function y = x p . (x p)" = p x p - 1	Constant y = C d(C) = 0 dx Power function y = x p . d (x p) = p x p - 1 d x
(a x)" = a x ln a	The exponential function y = a x . d (a x) = a x ln α d x In particular, for a = e we have y = e x d (e x) = e x d x
log a x " = 1 x ln a	Logarithmic functions y = log a x . d (log a x) = d x x ln a In particular, for a = e we have y = ln x d (ln x) = d x x
trigonometric functions. sin x "= cos x (cos x)" = - sin x (t g x) "= 1 c o s 2 x (c t g x)" = - 1 sin 2 x	trigonometric functions. d sin x = cos x d x d (cos x) = - sin x d x d (t g x) = d x c o s 2 x d (c t g x) = - d x sin 2 x
a r c sin x " = 1 1 - x 2 a r c cos x " = - 1 1 - x 2 a r c t g x " = 1 1 + x 2 a r c c t g x " = - 1 1 + x 2	Inverse trigonometric functions. d a r c sin x = d x 1 - x 2 d a r c cos x = - d x 1 - x 2 d a r c t g x = d x 1 + x 2 d a r c c t g x = - d x 1 + x 2

Let's illustrate the above with an example. Find the indefinite integral of the power function f (x) = x p .

According to the table of differentials d (x p) = p x p - 1 d x. By the properties of the indefinite integral, we have ∫ d (x p) = ∫ p x p - 1 d x = p ∫ x p - 1 d x = x p + C . Therefore, ∫ x p - 1 d x = x p p + C p , p ≠ 0. The second notation is as follows: ∫ x p d x = x p + 1 p + 1 + C p + 1 = x p + 1 p + 1 + C 1 , p ≠ - 1 .

Let's take it equal to - 1, find the set of antiderivatives of the power function f (x) = x p: ∫ x p d x = ∫ x - 1 d x = ∫ d x x .

Now we need a table of differentials for the natural logarithm d (ln x) = d x x , x > 0 , hence ∫ d (ln x) = ∫ d x x = ln x . Therefore ∫ d x x = ln x , x > 0 .

Table of antiderivatives (indefinite integrals)

The left column of the table contains formulas that are called basic antiderivatives. The formulas in the right column are not basic, but can be used to find indefinite integrals. They can be checked by differentiation.

Direct integration

To perform direct integration, we will use tables of antiderivatives, the integration rules ∫ f (k x + b) d x = 1 k F (k x + b) + C , and the properties of indefinite integrals ∫ k f (x) d x = k ∫ f (x) d x ∫ (f (x) ± g (x)) d x = ∫ f (x) d x ± ∫ g (x) d x

The table of basic integrals and the properties of the integrals can only be used after a slight transformation of the integrand.

Example 1

Let us find the integral ∫ 3 sin x 2 + cos x 2 2 d x

Solution

We take out the coefficient 3 from under the integral sign:

∫ 3 sin x 2 + cos x 2 2 d x = 3 ∫ sin x 2 + cos x 2 2 d x

Using the trigonometry formulas, we transform the integrand:

3 ∫ sin x 2 + cos x 2 2 d x = 3 ∫ sin x 2 2 + 2 sin x 2 cos x 2 + cos x 2 2 d x = = 3 ∫ 1 + 2 sin x 2 cos x 2 d x = 3 ∫ 1 + sin x d x

Since the integral of the sum is equal to the sum of the integrals, then
3 ∫ 1 + sin x d x = 3 ∫ 1 d x + ∫ sin x d x

We use the data from the table of antiderivatives: 3 ∫ 1 d x + ∫ sin x d x \u003d 3 (1 x + C 1 - cos x + C 2) \u003d \u003d n u t 3 C 1 + C 2 \u003d C \u003d 3 x - 3 cos x + C

Answer:∫ 3 sin x 2 + cos x 2 2 d x = 3 x - 3 cos x + C .

Example 2

It is necessary to find the set of antiderivatives of the function f (x) = 2 3 4 x - 7 .

Solution

We use the table of antiderivatives for the exponential function: ∫ a x · d x = a x ln a + C . This means that ∫ 2 x · d x = 2 x ln 2 + C .

We use the integration rule ∫ f (k x + b) d x = 1 k F (k x + b) + C .

We get ∫ 2 3 4 x - 7 d x = 1 3 4 2 3 4 x - 7 ln 2 + C = 4 3 2 3 4 x - 7 ln 2 + C .

Answer: f (x) = 2 3 4 x - 7 = 4 3 2 3 4 x - 7 ln 2 + C

Using the table of antiderivatives, properties and the rule of integration, we can find a lot of indefinite integrals. This is possible in those cases when it is possible to transform the integrand.

To find the integral of the logarithm function, the tangent and cotangent functions, and a number of others, special methods are used, which we will consider in the "Basic Integration Methods" section.

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In an earlier material, the question of finding the derivative was considered and its various applications: calculating the slope of the tangent to the graph, solving optimization problems, studying functions for monotonicity and extrema. $\newcommand(\tg)(\mathop(\mathrm(tg))\nolimits)$ $\newcommand(\ctg)(\mathop(\mathrm(ctg))\nolimits)$ $\newcommand(\arctg)( \mathop(\mathrm(arctg))\nolimits)$ $\newcommand(\arcctg)(\mathop(\mathrm(arctg))\nolimits)$

Picture 1.

The problem of finding the instantaneous speed $v(t)$ with the help of the derivative with respect to a previously known distance traveled, expressed by the function $s(t)$, was also considered.

Figure 2.

The inverse problem is also very common, when you need to find the path $s(t)$ traveled by a point in time $t$, knowing the speed of the point $v(t)$. If you remember, the instantaneous speed $v(t)$ is found as a derivative of the path function $s(t)$: $v(t)=s’(t)$. This means that in order to solve the inverse problem, that is, to calculate the path, you need to find a function whose derivative will be equal to the speed function. But we know that the derivative of the path is the speed, that is: $s'(t) = v(t)$. The speed is equal to the product of acceleration and time: $v=at$. It is easy to determine that the desired path function will have the form: $s(t) = \frac(at^2)(2)$. But this is not quite a complete solution. The complete solution will look like: $s(t)= \frac(at^2)(2)+C$, where $C$ is some constant. Why this is so will be discussed later. In the meantime, let's check the correctness of the found solution: $s"(t)=\left(\frac(at^2)(2)+C\right)"=2\frac(at)(2)+0=at=v( t)$.

It is worth noting that finding the path by speed is the physical meaning of the antiderivative.

The resulting function $s(t)$ is called the antiderivative of $v(t)$. Quite an interesting and unusual name, isn't it. There is a great meaning in it, which explains the essence of this concept and leads to its understanding. It can be seen that it contains two words "first" and "image". They speak for themselves. That is, this is the function that is the original for the derivative we have. And by this derivative we are looking for the function that was at the beginning, was the “first”, “first image”, that is, the antiderivative. It is sometimes also called a primitive function or an anti-derivative.

As we already know, the process of finding the derivative is called differentiation. And the process of finding the antiderivative is called integration. The integration operation is the inverse of the differentiation operation. The converse is also true.

Definition. An antiderivative for a function $f(x)$ on some interval is a function $F(x)$ whose derivative is equal to this function $f(x)$ for all $x$ from the specified interval: $F'(x)=f (x)$.

Someone may have a question: where did $F(x)$ and $f(x)$ come from in the definition, if initially it was about $s(t)$ and $v(t)$. The point is that $s(t)$ and $v(t)$ are special cases of notation for functions that have in this case specific meaning, that is, it is a function of time and a function of speed, respectively. The same is true for the variable $t$ - it represents the time. And $f$ and $x$ are the traditional variant of the general designation of a function and a variable, respectively. It is worth paying special attention to the notation of the antiderivative $F(x)$. First, $F$ is capital. Primitives are indicated by capital letters. Second, the letters are the same: $F$ and $f$. That is, for the function $g(x)$ the antiderivative will be denoted by $G(x)$, for $z(x)$ - by $Z(x)$. Regardless of the notation, the rules for finding the antiderivative function are always the same.

Let's look at a few examples.

Example 1 Prove that the function $F(x)=\frac(1)(5)\sin5x$ is the antiderivative of the function $f(x)=\cos5x$.

To prove this, we use the definition, or rather the fact that $F'(x)=f(x)$, and find the derivative of the function $F(x)$: $F'(x)=(\frac(1)(5 ) \sin5x)'=\frac(1)(5)\cdot 5\cos5x= \cos5x$. So $F(x)=\frac(1)(5) \sin5x$ is the antiderivative of $f(x)=\cos5x$. Q.E.D.

Example 2 Find to which functions the following antiderivatives correspond: a) $F(z)=\tg z$; b) $G(l) = \sin l$.

To find the desired functions, we calculate their derivatives:
a) $F'(z)=(\tg z)'=\frac(1)(\cos^2 z)$;
b) $G(l) = (\sin l)' = \cos l$.

Example 3 What will be the antiderivative for $f(x)=0$?
Let's use the definition. Let's think about which function can have a derivative equal to $0$. Remembering the table of derivatives, we get that any constant will have such a derivative. We get that the antiderivative we are looking for: $F(x)= C$.

The resulting solution can be explained geometrically and physically. Geometrically, it means that the tangent to the graph $y=F(x)$ is horizontal at each point of this graph and, therefore, coincides with the axis $Ox$. Physically explained by the fact that a point with a speed equal to zero remains in place, that is, the path traveled by it is unchanged. Based on this, we can formulate the following theorem.

Theorem. (Function constancy sign). If $F'(x) = 0$ on some interval, then the function $F(x)$ is constant on this interval.

Example 4 Determine the antiderivatives of which functions are the functions a) $F_1 = \frac(x^7)(7)$; b) $F_2 = \frac(x^7)(7) – 3$; c) $F_3 = \frac(x^7)(7) + 9$; d) $F_4 = \frac(x^7)(7) + a$, where $a$ is some number.
Using the definition of an antiderivative, we conclude that in order to solve this task, we need to calculate the derivatives of the antiderivative functions given to us. When calculating, remember that the derivative of a constant, that is, any number, is equal to zero.
a) $F_1 =(\frac(x^7)(7))"= 7 \cdot \frac(x^6)(7) = x^6$;
b) $F_2 =\left(\frac(x^7)(7) – 3\right)"=7 \cdot \frac(x^6)(7)= x^6$;
c) $F_3 =(\frac(x^7)(7) + 9)’= x^6$;
d) $F_4 =(\frac(x^7)(7) + a)’ = x^6$.

What do we see? Several different functions are antiderivatives of the same function. This means that any function has infinitely many antiderivatives, and they have the form $F(x) + C$, where $C$ is an arbitrary constant. That is, the operation of integration is multi-valued, in contrast to the operation of differentiation. Based on this, we formulate a theorem describing the main property of antiderivatives.

Theorem. (The main property of primitives). Let the functions $F_1$ and $F_2$ be antiderivatives of the function $f(x)$ on some interval. Then the following equality holds true for all values ​​from this interval: $F_2=F_1+C$, where $C$ is some constant.

The fact of the existence of an infinite set of antiderivatives can be interpreted geometrically. With the help of parallel translation along the axis $Oy$ one can obtain graphs of any two antiderivatives for $f(x)$ from each other. This is the geometric meaning of the antiderivative.

It is very important to pay attention to the fact that by choosing the constant $C$ it is possible to make the graph of the antiderivative pass through a certain point.

Figure 3

Example 5 Find the antiderivative for the function $f(x)=\frac(x^2)(3)+1$ whose graph passes through the point $(3; 1)$.
Let us first find all antiderivatives for $f(x)$: $F(x)=\frac(x^3)(9)+x + C$.
Next, we find a number C for which the graph $y=\frac(x^3)(9)+x + C$ will pass through the point $(3; 1)$. To do this, we substitute the coordinates of the point into the equation of the graph and solve it with respect to $C$:
$1= \frac(3^3)(9)+3 + C$, $C=-5$.
We got the graph $y=\frac(x^3)(9)+x-5$, which corresponds to the antiderivative $F(x)=\frac(x^3)(9)+x-5$.

Table of antiderivatives

A table of formulas for finding antiderivatives can be compiled using formulas for finding derivatives.

Table of antiderivatives

Functions	antiderivatives
$0$	$C$
$1$	$x+C$
$a\in R$	$ax+C$
$x^n, n\ne1$	$\displaystyle \frac(x^(n+1))(n+1)+C$
$\displaystyle \frac(1)(x)$	$\ln|x|+C$
$\sinx$	$-\cosx+C$
$\cos x$	$\sinx+C$
$\displaystyle \frac(1)(\sin^2 x)$	$-\ctgx+C$
$\displaystyle \frac(1)(\cos^2 x)$	$\tgx+C$
$e^x$	$e^x+C$
$a^x, a>0, a\ne1$	$\displaystyle \frac(a^x)(\ln a) +C$
$\displaystyle \frac(1)(\sqrt(1-x^2))$	$\arcsin x+C$
$\displaystyle -\frac(1)(\sqrt(1-x^2))$	$\arccos x+C$
$\displaystyle \frac(1)(1+x^2)$	$\arctgx+C$
$\displaystyle -\frac(1)(1+x^2)$	$\arctg x+C$

You can check the correctness of the table as follows: for each set of antiderivatives located in the right column, find the derivative, as a result of which the corresponding functions in the left column will be obtained.

Some rules for finding antiderivatives

As you know, many functions have a more complex form than those indicated in the table of antiderivatives, and can be any arbitrary combination of sums and products of functions from this table. And here the question arises, how to calculate the antiderivatives of similar functions. For example, from the table we know how to compute the antiderivatives $x^3$, $\sin x$ and $10$. But how, for example, to calculate the antiderivative $x^3-10\sin x$? Looking ahead, it is worth noting that it will be equal to $\frac(x^4)(4)+10\cos x$.
1. If $F(x)$ is an antiderivative for $f(x)$, $G(x)$ is for $g(x)$, then for $f(x)+g(x)$ the antiderivative will be equal to $ F(x)+G(x)$.
2. If $F(x)$ is an antiderivative for $f(x)$ and $a$ is a constant, then for $af(x)$ the antiderivative is $aF(x)$.
3. If for $f(x)$ the antiderivative is $F(x)$, $a$ and $b$ are constants, then $\frac(1)(a) F(ax+b)$ is antiderivative for $f (ax+b)$.
Using the obtained rules, we can expand the table of antiderivatives.

Functions	antiderivatives
$(ax+b)^n, n\ne1, a\ne0$	$\displaystyle \frac((ax+b)^n)(a(n+1)) +C$
$\displaystyle \frac(1)(ax+b), a\ne0$	$\displaystyle \frac(1)(a)\ln|ax+b|+C$
$e^(ax+b), a\ne0$	$\displaystyle \frac(1)(a) e^(ax+b)+C$
$\sin(ax+b), a\ne0$	$\displaystyle -\frac(1)(a)\cos(ax+b)+C$
$\cos(ax+b), a\ne0$	$\displaystyle \frac(1)(a)\sin(ax+b)+C$

Example 5 Find antiderivatives for:

a) $\displaystyle 4x^3+10x^7$;

b) $\displaystyle \frac(6)(x^5) -\frac(2)(x)$;

c) $\displaystyle 5\cos x+\sin(3x+15)$;

d) $\displaystyle \sqrt(x)-2\sqrt(x)$.

a) $4\frac (x^(3+1))(3+1)+10\frac(x^(7+1))(7+1)+C=x^4+\frac(5)( 4) x^8+C$;

b) $-\frac(3)(2x^4) -2\ln|x|+C$;

c) $5 \sin x - \frac(1)(3)\cos(3x + 15) + C$;

d) $\frac(2)(3)x\sqrt(x) - \frac(3)(2) x\sqrt(x) + C$.

Table of antiderivatives ("integrals"). Table of integrals. Tabular indefinite integrals. (Simple integrals and integrals with a parameter). Formulas for integration by parts. Newton-Leibniz formula.

Table of antiderivatives ("integrals"). Tabular indefinite integrals. (Simple integrals and integrals with a parameter).
Power function integral.	Power function integral.
An integral that reduces to an integral of a power function if x is driven under the sign of the differential.
	The exponential integral, where a is a constant number.
Integral of a compound exponential function.	The integral of the exponential function.
An integral equal to the natural logarithm.	Integral: "Long logarithm".
	Integral: "Long logarithm".
Integral: "High logarithm".	The integral, where x in the numerator is brought under the sign of the differential (the constant under the sign can be both added and subtracted), as a result, is similar to the integral equal to the natural logarithm.
Integral: "High logarithm".
Cosine integral.	Sine integral.
An integral equal to the tangent.	An integral equal to the cotangent.
Integral equal to both arcsine and arcsine
An integral equal to both the inverse sine and the inverse cosine.	An integral equal to both the arc tangent and the arc cotangent.
The integral is equal to the cosecant.	Integral equal to secant.
An integral equal to the arcsecant.	An integral equal to the arc cosecant.
An integral equal to the arcsecant.	An integral equal to the arcsecant.
An integral equal to the hyperbolic sine.	An integral equal to the hyperbolic cosine.
An integral equal to the hyperbolic sine, where sinhx is the hyperbolic sine in English.	An integral equal to the hyperbolic cosine, where sinhx is the hyperbolic sine in the English version.
An integral equal to the hyperbolic tangent.	An integral equal to the hyperbolic cotangent.
An integral equal to the hyperbolic secant.	An integral equal to the hyperbolic cosecant.

Formulas for integration by parts. Integration rules.

Formulas for integration by parts. Newton-Leibniz formula. Integration rules.
Integration of a product (function) by a constant:
Integration of the sum of functions:
indefinite integrals:
Integration by parts formula definite integrals:
Newton-Leibniz formula definite integrals:	Where F(a),F(b) are the values ​​of the antiderivatives at the points b and a, respectively.

Derivative table. Table derivatives. Derivative of the product. Derivative of private. Derivative of a complex function.

If x is an independent variable, then:

Derivative table. Table derivatives. "table derivative" - ​​yes, unfortunately, that's how they are searched on the Internet
	Power function derivative
	Derivative of the exponent
Derivative of a compound exponential function	Derivative of exponential function
Derivative of a logarithmic function	Derivative of the natural logarithm
	Derivative of the natural logarithm of a function
Sine derivative	cosine derivative
Cosecant derivative	Secant derivative
Derivative of arcsine	Arc cosine derivative
Derivative of arcsine	Arc cosine derivative
Tangent derivative	Cotangent derivative
Arc tangent derivative	Derivative of inverse tangent
Arc tangent derivative	Derivative of inverse tangent
Arcsecant derivative	Derivative of arc cosecant
Arcsecant derivative	Derivative of arc cosecant
Derivative of the hyperbolic sine Derivative of the hyperbolic sine in the English version	Hyperbolic cosine derivative The derivative of the hyperbolic cosine in the English version
Derivative of the hyperbolic tangent	Derivative of the hyperbolic cotangent
Derivative of hyperbolic secant	Derivative of the hyperbolic cosecant

Differentiation rules. Derivative of the product. Derivative of private. Derivative of a complex function.
Derivative of a product (function) by a constant:
Derivative of the sum (functions):
Derivative of the product (of functions):
The derivative of the quotient (of functions):
Derivative of a complex function:

Properties of logarithms. Basic formulas of logarithms. Decimal (lg) and natural logarithms (ln).

Basic logarithmic identity
Let us show how any function of the form a b can be made exponential. Since a function of the form e x is called exponential, then
Any function of the form a b can be represented as a power of ten

Natural logarithm ln (logarithm base e = 2.718281828459045…) ln(e)=1; log(1)=0

Taylor series. Expansion of a function in a Taylor series.

It turns out that most practically occurring mathematical functions can be represented with any accuracy in the vicinity of a certain point in the form of power series containing the powers of the variable in ascending order. For example, in the vicinity of the point x=1:

When using rows called taylor rows, mixed functions containing, say, algebraic, trigonometric, and exponential functions can be expressed as purely algebraic functions. With the help of series, differentiation and integration can often be quickly carried out.

The Taylor series in the vicinity of the point a has the following forms:

1) , where f(x) is a function that has derivatives of all orders at x=a. R n - the remainder term in the Taylor series is determined by the expression

2)

k-th coefficient (at x k) of the series is determined by the formula

3) A special case of the Taylor series is the Maclaurin series (=McLaren) (the decomposition takes place around the point a=0)

for a=0

the members of the series are determined by the formula

Conditions for the application of Taylor series.

1. In order for the function f(x) to be expanded in a Taylor series on the interval (-R;R), it is necessary and sufficient that the remainder term in the Taylor formula (Maclaurin (=McLaren)) for this function tends to zero at k →∞ on the specified interval (-R;R).

2. It is necessary that there are derivatives for this function at the point in the vicinity of which we are going to build a Taylor series.

Properties of Taylor series.

If f is an analytic function, then its Taylor series at any point a of the domain of f converges to f in some neighborhood of a.

There are infinitely differentiable functions whose Taylor series converges but differs from the function in any neighborhood of a. For example:

Taylor series are used for approximation (an approximation is a scientific method that consists in replacing some objects with others, in one sense or another close to the original, but simpler) functions by polynomials. In particular, linearization ((from linearis - linear), one of the methods of approximate representation of closed nonlinear systems, in which the study of a nonlinear system is replaced by an analysis of a linear system, in a sense equivalent to the original one.) of equations occurs by expanding into a Taylor series and cutting off all the terms above first order.

Thus, almost any function can be represented as a polynomial with a given accuracy.

Examples of some common expansions of power functions in Maclaurin series (=McLaren,Taylor in the vicinity of point 0) and Taylor in the vicinity of point 1. The first terms of expansions of the main functions in Taylor and MacLaren series.

Examples of some common expansions of power functions in Maclaurin series (= MacLaren, Taylor in the vicinity of the point 0)

Examples of some common Taylor series expansions around point 1

Principal Integrals Every Student Should Know

The listed integrals are the basis, the basis of the foundations. These formulas, of course, should be remembered. When calculating more complex integrals, you will have to use them constantly.

Pay special attention to formulas (5), (7), (9), (12), (13), (17) and (19). Do not forget to add an arbitrary constant C to the answer when integrating!

Integral of a constant

∫ A d x = A x + C (1)

Power function integration

In fact, one could confine oneself to formulas (5) and (7), but the rest of the integrals from this group are so common that it is worth paying a little attention to them.

∫ x d x = x 2 2 + C (2)
∫ x 2 d x = x 3 3 + C (3)
∫ 1 x d x = 2 x + C (4)
∫ 1 x d x = log | x | +C(5)
∫ 1 x 2 d x = − 1 x + C (6)
∫ x n d x = x n + 1 n + 1 + C (n ≠ − 1) (7)

Integrals of the exponential function and of hyperbolic functions

Of course, formula (8) (perhaps the most convenient to remember) can be considered as a special case of formula (9). Formulas (10) and (11) for the integrals of the hyperbolic sine and hyperbolic cosine are easily derived from formula (8), but it is better to just remember these relationships.

∫ e x d x = e x + C (8)
∫ a x d x = a x ln a + C (a > 0, a ≠ 1) (9)
∫ s h x d x = c h x + C (10)
∫ c h x d x = s h x + C (11)

Basic integrals of trigonometric functions

A mistake that students often make: they confuse the signs in formulas (12) and (13). Remembering that the derivative of the sine is equal to the cosine, for some reason many people believe that the integral of the sinx function is equal to cosx. This is not true! The integral of sine is "minus cosine", but the integral of cosx is "just sine":

∫ sin x d x = − cos x + C (12)
∫ cos x d x = sin x + C (13)
∫ 1 cos 2 x d x = t g x + C (14)
∫ 1 sin 2 x d x = − c t g x + C (15)

Integrals Reducing to Inverse Trigonometric Functions

Formula (16), which leads to the arc tangent, is naturally a special case of formula (17) for a=1. Similarly, (18) is a special case of (19).

∫ 1 1 + x 2 d x = a r c t g x + C = − a r c c t g x + C (16)
∫ 1 x 2 + a 2 = 1 a a r c t g x a + C (a ≠ 0) (17)
∫ 1 1 − x 2 d x = arcsin x + C = − arccos x + C (18)
∫ 1 a 2 − x 2 d x = arcsin x a + C = − arccos x a + C (a > 0) (19)

More complex integrals

These formulas are also desirable to remember. They are also used quite often, and their output is quite tedious.

∫ 1 x 2 + a 2 d x = ln | x + x2 + a2 | +C(20)
∫ 1 x 2 − a 2 d x = ln | x + x 2 − a 2 | +C(21)
∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 arcsin x a + C (a > 0) (22)
∫ x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 ln | x + x2 + a2 | + C (a > 0) (23)
∫ x 2 − a 2 d x = x 2 x 2 − a 2 − a 2 2 ln | x + x 2 − a 2 | + C (a > 0) (24)

General integration rules

1) The integral of the sum of two functions is equal to the sum of the corresponding integrals: ∫ (f (x) + g (x)) d x = ∫ f (x) d x + ∫ g (x) d x (25)

2) The integral of the difference of two functions is equal to the difference of the corresponding integrals: ∫ (f (x) − g (x)) d x = ∫ f (x) d x − ∫ g (x) d x (26)

3) The constant can be taken out of the integral sign: ∫ C f (x) d x = C ∫ f (x) d x (27)

It is easy to see that property (26) is simply a combination of properties (25) and (27).

4) Integral of a complex function if the inner function is linear: ∫ f (A x + B) d x = 1 A F (A x + B) + C (A ≠ 0) (28)

Here F(x) is the antiderivative for the function f(x). Note that this formula only works when the inner function is Ax + B.

Important: there is no universal formula for the integral of the product of two functions, as well as for the integral of a fraction:

∫ f (x) g (x) d x = ? ∫ f (x) g (x) d x = ? (thirty)

This does not mean, of course, that a fraction or a product cannot be integrated. It's just that every time you see an integral like (30), you have to invent a way to "fight" with it. In some cases, integration by parts will help you, somewhere you will have to make a change of variable, and sometimes even "school" formulas of algebra or trigonometry can help.

A simple example for calculating the indefinite integral

Example 1. Find the integral: ∫ (3 x 2 + 2 sin x − 7 e x + 12) d x

We use formulas (25) and (26) (the integral of the sum or difference of functions is equal to the sum or difference of the corresponding integrals. We get: ∫ 3 x 2 d x + ∫ 2 sin x d x − ∫ 7 e x d x + ∫ 12 d x

Recall that the constant can be taken out of the integral sign (formula (27)). The expression is converted to the form

3 ∫ x 2 d x + 2 ∫ sin x d x − 7 ∫ e ​​x d x + 12 ∫ 1 d x

Now let's just use the table of basic integrals. We will need to apply formulas (3), (12), (8) and (1). Let's integrate the power function, sine, exponent and constant 1. Don't forget to add an arbitrary constant C at the end:

3 x 3 3 - 2 cos x - 7 e x + 12 x + C

After elementary transformations, we get the final answer:

X 3 − 2 cos x − 7 e x + 12 x + C

Test yourself with differentiation: take the derivative of the resulting function and make sure that it is equal to the original integrand.

Summary table of integrals

∫ A d x = A x + C

∫ x d x = x 2 2 + C

∫ x 2 d x = x 3 3 + C

∫ 1 x d x = 2 x + C

∫ 1 x d x = log | x | + C

∫ 1 x 2 d x = − 1 x + C

∫ x n d x = x n + 1 n + 1 + C (n ≠ − 1)

∫ e x d x = e x + C

∫ a x d x = a x ln a + C (a > 0, a ≠ 1)

∫ s h x d x = c h x + C

∫ c h x d x = s h x + C

∫ sin x d x = − cos x + C

∫ cos x d x = sin x + C

∫ 1 cos 2 x d x = t g x + C

∫ 1 sin 2 x d x = − c t g x + C

∫ 1 1 + x 2 d x = a r c t g x + C = − a r c c t g x + C

∫ 1 x 2 + a 2 = 1 a a r c t g x a + C (a ≠ 0)

∫ 1 1 − x 2 d x = arcsin x + C = − arccos x + C

∫ 1 a 2 − x 2 d x = arcsin x a + C = − arccos x a + C (a > 0)

∫ 1 x 2 + a 2 d x = ln | x + x2 + a2 | + C

∫ 1 x 2 − a 2 d x = ln | x + x 2 − a 2 | + C

∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 arcsin x a + C (a > 0)

∫ x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 ln | x + x2 + a2 | + C (a > 0)

∫ x 2 − a 2 d x = x 2 x 2 − a 2 − a 2 2 ln | x + x 2 − a 2 | + C (a > 0)

Download the table of integrals (part II) from this link

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