Table of uncertain integrals complete. Basic formulas and integration methods
In school, many are not able to solve the integrals or any difficulties arise with them. This article will help you to figure it out, because in it you will find everything Tables integrals.
Integralit is one of the main computing and concept in mathematical analysis. His appearance came from two purposes:
First goal - Restore the function with its derivative.
Second goal- Calculation of the area located at a distance from the graph to the function f (x) on the line where, and more or equal to x is greater than or equal to B and the abscissa axis.
These goals summarize us to certain and uncertain integrals. The connection between these integrals lies in the search for properties and calculation. But all the flows and everything changes over time, there were new solutions, there were thereby addresses certain and indefinite integrals to other integration forms.
What uncertain integral you ask. This is a primitive function f (x) of one variable x in the interval A greater x more b. called any function f (x) in this interval For any designation X, the derivative is equal to F (X). It is clear that f (x) is primitive for f (x) in the interval A greater x more b. So F1 (x) \u003d f (x) + C. C is any constant and primitive for F (X) in this interval. This statement is reversible, for the function f (x) - 2, first-shaped differ only on constant. Relying on the integrated calculus theorem It turns out that each continuous in the interval A
Certain integral it is understood as the limit in integral sums, or in a situation of a given function f (x), a typical F, which means the difference of its expressions on the ends of this direct F (B) - F (a).
For visibility to study this topic, I suggest watching a video. It tells in detail and shows how to find integrals.
Each integral table itself is very useful, as it helps in solving a specific type of integrals.
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The main integrals that each student should know
The listed integrals are the basis, the foundation base. These formulas should be remembered. When calculating more complex integrals You will have to use them constantly.
Note special attention To formulas (5), (7), (9), (12), (13), (17) and (19). Do not forget when integrating add to the answer arbitrary constant with!
Integral from Constanta
∫ a d x \u003d a x + c (1)Integrating power function
In fact, it was possible to limit only by formulas (5) and (7), but the rest of the integrals from this group are encountered as often that it is worth paying a little attention to them.
∫ x D x \u003d x 2 2 + C (2)
∫ x 2 d x \u003d x 3 3 + c (3)
∫ 1 x D x \u003d 2 x + c (4)
∫ 1 x D x \u003d ln | X | + C (5)
∫ 1 x 2 D x \u003d - 1 x + c (6)
∫ x n d x \u003d x n + 1 n + 1 + c (n ≠ - 1) (7)
Integrals from the indicative function and from hyperbolic functions
Of course, formula (8) (perhaps the most convenient for remembrance) can be considered as a special case of formula (9). Formulas (10) and (11) for integrals from hyperbolic sinus and hyperbolic cosine are easily derived from formula (8), but it is better to just remember these relations.
∫ e x d x \u003d e x + c (8)
∫ a x d x \u003d a x ln a + c (a\u003e 0, a ≠ 1) (9)
∫ s H x D x \u003d c h x + c (10)
∫ C H x D x \u003d s h x + c (11)
Basic integrals from trigonometric functions
An error that students often makes: confuse signs in Formulas (12) and (13). By remembering that the sinus derivative is equal to cosine, many for some reason consider that the integral from the SINX function is COSX. This is not true! The integral of sine is equal to "minus cosine", but the integral from COSX is "just sinus":
∫ SIN X D X \u003d - COS X + C (12)
∫ cos x d x \u003d sin x + c (13)
∫ 1 COS 2 x D X \u003d T G X + C (14)
∫ 1 SIN 2 x D X \u003d - C T G X + C (15)
Integrals reduced to inverse trigonometric functions
Formula (16), leading to Arctangent, naturally, is a special case of formula (17) at a \u003d 1. Similarly, (18) - a special case (19).
∫ 1 1 + x 2 d x \u003d a r c t g x + c \u003d - a r c c t g x + c (16)
∫ 1 x 2 + a 2 \u003d 1 A A R C T G X A + C (A ≠ 0) (17)
∫ 1 1 - x 2 D x \u003d Arcsin X + C \u003d - Arccos X + C (18)
∫ 1 A 2 - X 2 D X \u003d Arcsin X A + C \u003d - Arccos X A + C (A\u003e 0) (19)
More complex integrals
These formulas are also desirable to remember. They are also used quite often, and their conclusion is quite tedious.
∫ 1 x 2 + a 2 d x \u003d ln | x + x 2 + a 2 | + C (20)
∫ 1 x 2 - a 2 d x \u003d ln | X + X 2 - A 2 | + C (21)
∫ a 2 - x 2 D x \u003d x 2 a 2 - x 2 + a 2 2 ARCSIN X A + C (A\u003e 0) (22)
∫ x 2 + a 2 d x \u003d x 2 x 2 + a 2 + a 2 2 ln | x + x 2 + a 2 | + C (A\u003e 0) (23)
∫ x 2 - a 2 d x \u003d x 2 x 2 - a 2 - a 2 2 ln | X + X 2 - A 2 | + C (A\u003e 0) (24)
General integration rules
1) The integral from the sum of two functions is equal to the sum of the corresponding integrals: ∫ (f (x) + g (x)) D x \u003d ∫ f (x) d x + ∫ g (x) d x (25)
2) the integral of the difference of two functions is equal to the difference between the corresponding integrals: ∫ (F (x) - G (x)) D x \u003d ∫ 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 \u003d c ∫ f (x) d x (27)
It is easy to notice that property (26) is just a combination of properties (25) and (27).
4) integral from a complex function if the internal function is linear: ∫ f (a x + b) d x \u003d 1 a f (a x + b) + c (a ≠ 0) (28)
Here f (x) is a primitive for the function f (x). Note: This formula is suitable only for the case when an internal function has a view AX + B.
Important: There is no universal formula for integral from the product of two functions, as well as for the integral from the fraction:
∫ f (x) g (x) d x \u003d? ∫ f (x) g (x) d x \u003d? (thirty)
This does not mean, of course, that the fraction or the work cannot be integrated. Just every time, seeing an integral type (30), you will have to invent the way "fighting" with him. In some cases, the integration of parts will help you, somewhere will have to replace the variable, and sometimes assistance can even provide "school" formulas algebra or trigonometry.
A simple example of calculating an uncertain integral
Example 1. Find an integral: ∫ (3 x 2 + 2 SIN x - 7 E x + 12) D xWe use the formulas (25) and (26) (the integral of the amount or difference of functions is equal to the sum or the difference of the corresponding integrals. We obtain: ∫ 3 x 2 D x + ∫ 2 sin x D x - ∫ 7 E x D x + ∫ 12 D x
Recall that the constant can be made over the integral sign (formula (27)). Expression converted to mind
3 ∫ x 2 d x + 2 ∫ sin x d x - 7 ∫ e \u200b\u200bx d x + 12 ∫ 1 d x
And now simply use the table of the main integrals. We will need to apply formulas (3), (12), (8) and (1). Integrated power function, sinus, exponent and constant 1. Do not forget add to the end of an arbitrary constant with:
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
Check yourself with differentiation: Take the derivative of the obtained function and make sure that it is equal to the initial election expression.
Summary integral table
| ∫ a d x \u003d a x + c |
| ∫ x D x \u003d x 2 2 + C |
| ∫ x 2 d x \u003d x 3 3 + c |
| ∫ 1 x D x \u003d 2 x + c |
| ∫ 1 x D x \u003d ln | X | + C. |
| ∫ 1 x 2 d x \u003d - 1 x + c |
| ∫ x n d x \u003d x n + 1 n + 1 + c (n ≠ - 1) |
| ∫ E x D x \u003d e x + c |
| ∫ a x d x \u003d a x ln a + c (a\u003e 0, a ≠ 1) |
| ∫ s h x d x \u003d c h x + c |
| ∫ C h x d x \u003d s h x + c |
| ∫ sin x d x \u003d - cos x + c |
| ∫ cos x d x \u003d sin x + c |
| ∫ 1 COS 2 x D X \u003d T G X + C |
| ∫ 1 SIN 2 x D X \u003d - C T G X + C |
| ∫ 1 1 + x 2 d x \u003d a r c t g x + c \u003d - a r c c t g x + c |
| ∫ 1 x 2 + a 2 \u003d 1 A A R C T G X A + C (A ≠ 0) |
| ∫ 1 1 - x 2 d x \u003d arcsin x + c \u003d - arccos x + c |
| ∫ 1 A 2 - X 2 D X \u003d Arcsin X A + C \u003d - Arccos X A + C (A\u003e 0) |
| ∫ 1 x 2 + a 2 d x \u003d ln | x + x 2 + a 2 | + C. |
| ∫ 1 x 2 - a 2 d x \u003d ln | X + X 2 - A 2 | + C. |
| ∫ a 2 - x 2 d x \u003d x 2 a 2 - x 2 + a 2 2 arcsin x a + c (a\u003e 0) |
| ∫ x 2 + a 2 d x \u003d x 2 x 2 + a 2 + a 2 2 ln | x + x 2 + a 2 | + C (A\u003e 0) |
| ∫ x 2 - a 2 d x \u003d x 2 x 2 - a 2 - a 2 2 ln | X + X 2 - A 2 | + C (A\u003e 0) |
Download the integral table (part II) on this link
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We list the integrals from the elementary functions that are sometimes called tables:
Any of the above formulas can be proved by taking a derived from the right side (as a result, a source function will be obtained).
Integration methods
Consider some basic integration methods. These include:
1. The decomposition method(direct integration).
This method is on the direct application of tabular integrals, as well as on the application of properties 4 and 5 of an indefinite integral (i.e., on the removal of a constant factory and / or representation of the integrand function in the form of the amount of functions - decomposition of the integrated function to the components).
Example 1.For example, for finding (dx / x 4), you can directly use the table integral forx N DX. In fact, (dx / x 4) \u003d x -4 dx \u003d x -3 / (- 3) + c \u003d -1 / 3x 3 + c.
Consider a few more examples.
Example 2.For a while using the same integral:
Example 3.For served in the championships
Example 4.To find, imagine a source function in the form of 
and we use a tabular integral for the indicative function:
Consider the use of the removal for the bracket of a constant in fact.
Example 5.
We will find, for example 
. Considering that, we get
Example 6.We will find. Insofar as 
, use the tabular integral 
Receive
In the following two examples, you can also use the removal of brackets and tabular integrals:
Example 7.
(Using I. 
);
Example 8.
(Using 
and 
).
Consider more complex examples that use the integral amount.
Example 9.For example, we will find 
. To apply the decomposition method in the numerator, use the formula of the cube of the amount , and then the polynomial obtained is mined to the denominator.
\u003d ((8x 3/2 + 12x + 6x 1/2 + 1) / (x 3/2)) dx \u003d (8 + 12x -1/2 + 6 / x + x -3/2) DX \u003d 8 DX + 12X -1/2 DX + 6DX / X + X -3/2 DX \u003d 
It should be noted that at the end of the decision, one overall permanent C (and not separate during the integration of each term) was recorded. In the future, it is also proposed to omit in the process of solutions constant on the integration of individual terms as long as the expression contains at least one indefinite integral (we will record one constant at the end of the decision).
Example 10.Find 
. To solve this task, we will decompose the numerator on multipliers (after that it will be possible to reduce the denominator).
Example 11.We will find. Here you can use trigonometric identities.
Sometimes, in order to decompose the expression on the components, you have to use more complex techniques.
Example 12.Find 
. In the integrated function, we allocate the whole part of the fraction 
. Then
Example 13.Find
2. Method of replacement of the variable (substitution method)
The method is based on the following formula: f (x) dx \u003d f ( (t)) (t) dt, where x \u003d (t) is a function differentiated on the interval under consideration.
Evidence. We find derivatives on the variable TOT left and right parts Formulas.
Note that in the left part there is a complex function, the intermediate argument of which is x \u003d (t). Therefore, in order to differentiate its PT, first differentiate the integral of X, and then the derivative of the PT intermediate argument.
( f (x) dx) `T \u003d ( f (x) dx)` x * x` T \u003d f (x) ` (t)
The derived from the right side:
(f ( (t)) (t) dt) `t \u003d f ( (t)) (t) \u003d f (x) (t)
Since these derivatives are equal, by consequence of the Lagrange Theorem, the left and right parts of the proved formula differ in some permanent. Since the indefinite integrals themselves are defined with an accuracy of an indefinite constant component, the indicated constant in the final record can be omitted. Proved.
Successful replacement variable allows you to simplify the original integral, and in the simplest cases to reduce it to the table. The use of this method distinguishes the methods of a linear and nonlinear substitution.
a) linear substitution methodconsider on the example.
Example 1.
. Let \u003d 1 - 2x, then
dX \u003d D (½ - ½T) \u003d - ½DT
It should be noted that the new variable can not be written explicitly. In such cases, they are talking about the transformation of the function under the sign of the differential or the introduction of constant and variables under the sign of the differential, i.e. about implicit replacement of the variable.
Example 2.For example, we findcos (3x + 2) DX. According to the properties of the differential DX \u003d (1/3) D (3x) \u003d (1/3) D (3x + 2), thencos (3x + 2) DX \u003d (1/3) COS (3x + 2) D (3x + + 2) \u003d (1/3) COS (3X + 2) D (3x + 2) \u003d (1/3) sin (3x + 2) + c.
In both considered examples, a linear substitution T \u003d KX + B (K0) was used to find the integrals.
In general, the following theorem is valid.
Linear substitution theorem. Let (x) be some primitive for the functionf (x). Then f (kx + b) dx \u003d (1 / k) f (kx + b) + C, where k and b are some permanent, K0.
Evidence.
By definition of the integral f (kx + b) d (kx + b) \u003d f (kx + b) + c. HOD (KX + B) \u003d (KX + B) `DX \u003d KDX. I will put a permanent multiplier sign of the integral sign: kf (kx + b) dx \u003d f (kx + b) + c. Now it is possible to divide the left and right parts of the equality of order to obtain a proven statement with an accuracy to the designation of the constant component.
This theorem argues that if in the definition of the integral f (x) dx \u003d f (x) + c instead of the argument x substitute the expression (kx + b), then this will lead to the appearance of an additional multiplier 1 / kpreted primary.
Using the proven theorem will solve the following examples.
Example 3.
Find 
. Herekx + b \u003d 3 -X, i.e.k \u003d -1, b \u003d 3. then
Example 4.
We will find. Herekx + b \u003d 4x + 3, i.e.k \u003d 4, b \u003d 3. then
Example 5.
Find 
. Herekx + b \u003d -2x + 7, i.e.k \u003d -2, b \u003d 7. Then
.
Example 6.Find 
. Herekx + b \u003d 2x + 0, i.e.k \u003d 2, b \u003d 0.
.
Compare the result obtained with Example 8, which was resolved by the decomposition method. Solving the same task to another method, we got the answer 
. Compare the results obtained:. Thus, these expressions differ from each other for a permanent term 
. The received answers do not contradict each other.
Example 7.Find 
. We highlight the full square in the denominator.
In some cases, the replacement of the variable does not reduce the integral directly to the table, but can simplify the solution by making it possible to use in the subsequent step of the decomposition method.
Example 8.For example, we will find 
. Replace \u003d X + 2, thenDT \u003d D (X + 2) \u003d DX. Then
,
where C \u003d C 1 - 6 (when substituting instead of exploration (X + 2), instead of the first two components, we obtain ½x 2 -2x- 6).
Example 9.Find 
. Let \u003d 2x + 1, thenDT \u003d 2DX; DX \u003d ½DT; X \u003d (T- 1) / 2.
We substitute instead of tension (2x + 1), open brackets and give the like.
Note that in the process of transformations, we switched to another constant term, because The group of permanent terms in the process of transformations could be omitted.
b) nonlinear substitution methodconsider on the example.
Example 1.
. Let \u003d -x 2. Next, it would be possible to express x middle, then find the expression for the DX and implement the replacement of the variable in the search integral. But in this case it is easier to do differently. We finddt \u003d d (-x 2) \u003d -2xdx. It should be noted that the expressionXDX is the fact about the integrand expression of the desired integral. Express it from the obtained equalityXDX \u003d - ½DT. Then
The following four main integration methods are listed.
1)
The integration rule of the amount or difference.
.
Hereinafter, u, V, W - functions from the integration variable x.
2)
Making a permanent integral sign.
Let c be a constant, independent of x. Then it can be taken out for the sign of the integral.
3)
Method of replacement variable.
Consider an indefinite integral.
If you manage to pick up such a function φ (x) from x so
,
By replacing the variable T \u003d φ (x), we have
.
4)
Integration formula in parts.
,
where u and v are functions from integration variable.
The ultimate goal of calculating uncertain integrals is, by transformations, bring the specified integral to the simplest integrals called tables. Table integrals are expressed through elementary functions by famous formulas.
See Table Integrals \u003e\u003e\u003e
Example
Calculate an indefinite integral
Decision
We notice that the integrated function is the sum and the difference of three members:
, and.
Apply the method 1
.
Next, we notice that the integrand functions of new integrals are multiplied by permanent 5, 4,
and 2
, respectively. Apply the method 2
.
In the integral table we find the formula
.
Believing n \u003d 2
, we find the first integral.
Rewrite the second integral in the form
.
We notice that. Then
Apply the third method. We make the replacement of the variable T \u003d φ (x) \u003d ln x.
.
In the integral table we find the formula
Since the integration variable can be denoted by any letter,
Rewrite the third integral in the form
.
Apply the integration formula in parts.
Put.
Then
;
;
;
;
.
Finally have
.
We collect members with x 3
.
.
Answer
References:
N.M. Gunter, R.O. Kuzmin, Collection of tasks higher Mathematics, "Lan", 2003.
Primed function and indefinite integral
Fact 1. Integration - action, inverse differentiation, namely, restoring the function according to a known derivative of this function. Function restored F.(x.) Called predo-shaped For function f.(x.).
Definition 1. Function F.(x. f.(x.) at some interval X.if for all values x. equality is performed from this gap F. "(x.)=f.(x.), that is, this feature f.(x.) is a derivative of a primitive function F.(x.). .
For example, a function F.(x.) \u003d sin. x. is a primary for function f.(x.) \u003d COS. x. on the whole numerical straight, since with any value of the IKSA (sin. x.) "\u003d (COS x.) .
Definition 2. Uncertainly integral function f.(x.) It is called the totality of all its primitive. This uses recording
∫
f.(x.)dX.
,where sign ∫ called the integral sign, function f.(x.) - a replacement function, and f.(x.)dX. - A concrete expression.
Thus, if F.(x.) - some kind of primary for f.(x.), T.
∫
f.(x.)dX. = F.(x.) +C.
where C. - arbitrary constant (constant).
To understand the meaning of many primitive functions as an indefinite integral, the following analogy is appropriate. Let there be a door (traditional wooden door). Its function is "to be a door." And what is the door made from? From wood. Therefore, a multitude of primitive integrated function "Be the door", that is, it is an indefinite integral, is the function "Being + C", where C is a constant, which in this context may indicate, for example, a tree of wood. Just as the door is made of wood using some tools, the derivative of the "made" function from the primitive function with the formulas that we learned by studying the derivative .
Then the table of the functions of common objects and the corresponding primitive ("to be the door" - "be tree", "be a spoon" - "be metal", etc.) is similar to the table of the main indefinite integrals, which will be shown slightly below. The table of uncertain integrals lists common functions with the indication of the primordial, of which these functions are made. In terms of the tasks to find a indefinite integral, such integrants are given, which without particular gravity can be integrated directly, that is, on the table of uncertain integrals. In the tasks, it is necessary to pre-convert to the tasks to preform so that you can use table integrals.
Fact 2. Restoring the function as a primitive, we must take into account an arbitrary constant (constant) C., so as not to write a list of primitive with different constants from 1 to infinity, you need to record many of the primitive with an arbitrary constant C.For example, as follows: 5 x.³ + s. So, an arbitrary constant (constant) enters the expression of primitive, since the primitive can be a function, for example, 5 x.³ + 4 or 5 x.³ + 3 and with differentiation 4 or 3, or any other constant is applied to zero.
We will put the integration task: for this function f.(x.) find such a function F.(x.), derivative of which equal f.(x.).
Example 1.Find a variety of features
Decision. For this feature, the function is function
Function F.(x.) called primitive for function f.(x.) if derivative F.(x.) Equal f.(x.), or that the same, differential F.(x.) Raven f.(x.) dX..
(2)
Consequently, the function is primitive for a function. However, it is not the only primary for. They also serve as functions
where FROM - Arbitrary constant. This can be seen differentiation.
Thus, if there is one first primary for the function, then it has an infinite multitude of primitive, differing in permanent term. All the primary functions are written in the above form. This follows from the following theorem.
Theorem (formal statement of fact 2).If a F.(x.) - Valid for function f.(x.) at some interval H., then any other primitive for f.(x.) At the same gap can be presented in the form F.(x.) + C.where FROM- Arbitrary constant.
In the following example, we already appeal to the integral table, which will be given in paragraph 3, after the properties of an indefinite integral. We do it before familiarization with the entire table, so that the essence of the foregoing is understood. And after the table and properties we will use them when integrating in all fullness.
Example 2.Find multiple features:
Decision. We find the sets of primitive functions, of which "these functions are made". When mentioning the formulas from the integral table, simply accept that there are such formulas, and we will study the table of uncertain integrals to be completely further.
1) applying formula (7) from the integral table with n. \u003d 3, we get
2) using formula (10) from the integral table with n. \u003d 1/3, we have
3) as
then by formula (7) when n. \u003d -1/4 Find
Under the sign of the integral write not the function itself f. , and her work on differential dX. . This is done primarily in order to indicate which variable is looking for a primitive. For example,
,
;
here, in both cases, the integrand function is equal, but its indefinite integrals in the considered cases are different. In the first case, this feature is considered as a function from a variable x. , and in the second - as a function from z. .
The process of finding an indefinable integral function is called integrating this function.
Geometric meaning of an indefinite integral
Let it be required to find a curve y \u003d f (x) And we already know that tangent tilt angle at each point is set function f (x) The abscissions of this point.
According to the geometric meaning of the derivative, tangent tilt angle at this point of the curve y \u003d f (x) equal to the value of the derivative F "(x). So you need to find such a function F (x), for which F "(x) \u003d f (x). Function required in the task F (x) is a primary one f (x). The condition of the problem satisfies not one curve, but the family of curves. y \u003d f (x) - one of such curves, and every other curve can be obtained from her parallel transfer along the axis Oy..
Let's call a graph of a primitive function from f (x) integral curve. If a F "(x) \u003d f (x)then the graph of the function y \u003d f (x) There is an integral curve.
Fact 3. An uncertain integral is geometrically represented by the seven of all integrated curves As in the figure below. The remoteness of each curve from the start of coordinates is determined by an arbitrary constant (constant) integration C..
Properties of an indefinite integral
Fact 4. Theorem 1. The derivative of an indefinite integral is equal to the integrand function, and its differential is a source expression.
Fact 5. Theorem 2. Unexposed integral from differential function f.(x.) equal function f.(x.) with an accuracy of a permanent term .
(3)
Theorems 1 and 2 show that differentiation and integration are mutually reverse operations.
Fact 6. Theorem 3. A constant multiplier in the integrand can be made for a sign of an indefinite integral .
