Table of antiderivative mathematical functions. Antiderivative function and indefinite integral. Logarithmic functions y = log a x

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

Table of antiderivatives

We can find the antiderivative from a 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.

Let's 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 d x Power function y = x p. d (x p) = p x p - 1 d x
(a x) " = a x ln a	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 us illustrate the above with an example. Let's 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 version of the entry 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 us take it equal to - 1 and 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, therefore ∫ 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, integration rules ∫ f (k x + b) d x = 1 k F (k x + b) + C, as well as 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 properties of integrals can be used only after an easy transformation of the integrand.

Example 1

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

Solution

We remove 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 trigonometry formulas, we transform the integrand function:

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 = 3 (1 x + C 1 - cos x + C 2) = = empty 3 C 1 + C 2 = C = 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 cases where it is possible to transform the integrand.

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

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Let us list the integrals of elementary functions, which are sometimes called tabular:

Any of the above formulas can be proven by taking the derivative of the right-hand side (the result will be the integrand).

Integration methods

Let's look at some basic integration methods. These include:

1. Decomposition method(direct integration).

This method is based on the direct use of tabular integrals, as well as on the use of properties 4 and 5 of the indefinite integral (i.e., taking the constant factor out of brackets and/or representing the integrand as a sum of functions - decomposition of the integrand into terms).

Example 1. For example, to find(dx/x 4) you can directly use the table integral forx n dx. In fact,(dx/x 4) =x -4 dx=x -3 /(-3) +C= -1/3x 3 +C.

Example 2. To find it, we use the same integral:

Example 3. To find it you need to take

Example 4. To find, we represent the integrand function in the form and use the table integral for the exponential function:

Let's consider the use of bracketing a constant factor.

Example 5.Let's find, for example . Considering that, we get

Example 6. We'll find it. Because the , let's use the table integral We get

In the following two examples, you can also use bracketing and table integrals:

Example 7.

(we use and );

Example 8.

(we use And ).

Let's look at more complex examples that use the sum integral.

Example 9. For example, let's find
. To apply the expansion method in the numerator, we use the formula for the cube of the sum  , and then divide the resulting polynomial term by term by the denominator.

=((8x 3/2 + 12x+ 6x 1/2 + 1)/(x 3/2))dx=(8 + 12x -1/2 + 6/x+x -3/2)dx= 8 dx+ 12x -1/2 dx+ + 6dx/x+x -3/2 dx=

It should be noted that at the end of the solution one common constant C is written (and not separate ones when integrating each term). In the future, it is also proposed to omit the constants from the integration of individual terms in the solution process as long as the expression contains at least one indefinite integral (we will write one constant at the end of the solution).

Example 10. We'll find . To solve this problem, let's factorize the numerator (after this we can reduce the denominator).

Example 11. We'll find it. Trigonometric identities can be used here.

Sometimes, in order to decompose an expression into terms, you have to use more complex techniques.

Example 12. We'll find . In the integrand we select the whole part of the fraction . Then

Example 13. We'll find

2. Variable replacement method (substitution method)

The method is based on the following formula: f(x)dx=f((t))`(t)dt, where x =(t) is a function differentiable on the interval under consideration.

Proof. Let's find the derivatives with respect to the variable t from the left and right sides of the formula.

Note that on the left side there is a complex function whose intermediate argument is x = (t). Therefore, to differentiate it with respect to t, we first differentiate the integral with respect to x, and then take the derivative of the intermediate argument with respect to t.

( f(x)dx)` t = ( f(x)dx)` x *x` t = f(x) `(t)

Derivative from the right side:

(f((t))`(t)dt)` t =f((t))`(t) =f(x)`(t)

Since these derivatives are equal, by corollary to Lagrange’s theorem, the left and right sides of the formula being proved differ by a certain constant. Since the indefinite integrals themselves are defined up to an indefinite constant term, this constant can be omitted from the final notation. Proven.

A successful change of variable allows you to simplify the original integral, and in the simplest cases, reduce it to a tabular one. In the application of this method, a distinction is made between linear and nonlinear substitution methods.

a) Linear substitution method Let's look at an example.

Example 1.
. Let t= 1 – 2x, then

dx=d(½ - ½t) = - ½dt

It should be noted that the new variable does not need to be written out explicitly. In such cases, they talk about transforming a function under the differential sign or about introducing constants and variables under the differential sign, i.e. O implicit variable replacement.

Example 2. For example, let's findcos(3x + 2)dx. By the properties of the differential dx = (1/3)d(3x) = (1/3)d(3x + 2), thencos(3x + 2)dx =(1/3)cos(3x + 2)d (3x + + 2) = (1/3)cos(3x + 2)d(3x + 2) = (1/3)sin(3x + 2) +C.

In both examples considered, linear substitution t=kx+b(k0) was used to find the integrals.

In the general case, the following theorem is valid.

Linear substitution theorem. Let F(x) be some antiderivative of the function f(x). Thenf(kx+b)dx= (1/k)F(kx+b) +C, where k and b are some constants,k0.

Proof.

By definition of the integral f(kx+b)d(kx+b) =F(kx+b) +C. Hod(kx+b)= (kx+b)`dx=kdx. Let's take the constant factor k out of the integral sign: kf(kx+b)dx=F(kx+b) +C. Now we can divide the left and right sides of the equality into two and obtain the statement to be proved up to the designation of the constant term.

This theorem states that if in the definition of the integral f(x)dx= F(x) + C instead of the argument x we ​​substitute the expression (kx+b), this will lead to the appearance of an additional factor 1/k in front of the antiderivative.

Using the proven theorem, we solve the following examples.

Example 3.

We'll find . Here kx+b= 3 –x, i.e. k= -1,b= 3. Then

Example 4.

We'll find it. Herekx+b= 4x+ 3, i.e. k= 4,b= 3. Then

Example 5.

We'll find . Here kx+b= -2x+ 7, i.e. k= -2,b= 7. Then

.

Example 6. We'll find
. Here kx+b= 2x+ 0, i.e. k= 2,b= 0.

.

Let us compare the result obtained with example 8, which was solved by the decomposition method. Solving the same problem using a different method, we got the answer
. Let's compare the results: Thus, these expressions differ from each other by a constant term , i.e. The answers received do not contradict each other.

Example 7. We'll find
. Let's select a perfect square in the denominator.

In some cases, changing a variable does not reduce the integral directly to a tabular one, but can simplify the solution, making it possible to use the expansion method at a subsequent step.

Example 8. For example, let's find . Replace t=x+ 2, then dt=d(x+ 2) =dx. Then

,

where C = C 1 – 6 (when substituting the expression (x+ 2) instead of the first two terms we get ½x 2 -2x– 6).

Example 9. We'll find
. Let t= 2x+ 1, then dt= 2dx;dx= ½dt;x= (t– 1)/2.

Let's substitute the expression (2x+ 1) for t, open the brackets and give similar ones.

Note that in the process of transformations we moved to another constant term, because the group of constant terms could be omitted during the transformation process.

b) Nonlinear substitution method Let's look at an example.

Example 1.
. Lett= -x 2. Next, one could express x in terms of t, then find an expression for dx and implement a change of variable in the desired integral. But in this case it’s easier to do things differently. Let's finddt=d(-x 2) = -2xdx. Note that the expression xdx is a factor of the integrand of the desired integral. Let us express it from the resulting equalityxdx= - ½dt. Then

=  (- ½)e t dt = (- ½) e t dt = (- ½)e t + C = (- ½)
+C

Let's look at a few more examples.

Example 2. We'll find . Let t= 1 -x 2 . Then

Example 3. We'll find . Lett=. Then

;

Example 4. In the case of nonlinear substitution, it is also convenient to use implicit variable substitution.

For example, let's find
. Let's write xdx= = (-1/4)d(3 - 2x 2) (implicitly replaced by the variable t= 3 - 2x 2). Then

Example 5. We'll find . Here we also introduce a variable under the differential sign: (implicit replacement = 3 + 5x 3). Then

Example 6. We'll find . Because the ,

Example 7. We'll find it. Since then

Let's look at a few examples in which it becomes necessary to combine various substitutions.

Example 8. We'll find
. Lett= 2x+ 1, thenx= (t– 1)/2;dx= ½dt.

Example 9. We'll find
. Lett=x- 2, thenx=t+ 2;dx=dt.

Antiderivative function and indefinite integral

Fact 1. Integration is the inverse action of differentiation, namely, restoring a function from the known derivative of this function. The function thus restored F(x) is called antiderivative for function f(x).

Definition 1. Function F(x f(x) on some interval X, if for all values x from this interval the equality holds F "(x)=f(x), that is, this function f(x) is the derivative of the antiderivative function F(x). .

For example, the function F(x) = sin x is an antiderivative of the function f(x) = cos x on the entire number line, since for any value of x (sin x)" = (cos x) .

Definition 2. Indefinite integral of a function f(x) is the set of all its antiderivatives. In this case, the notation is used

∫

f(x)dx

,

where is the sign ∫ called the integral sign, the function f(x) – integrand function, and f(x)dx – integrand expression.

Thus, if F(x) – some antiderivative for f(x) , That

∫

f(x)dx = F(x) +C

Where C - arbitrary constant (constant).

To understand the meaning of the set of antiderivatives of a function as an indefinite integral, the following analogy is appropriate. Let there be a door (traditional wooden door). Its function is to “be a door.” What is the door made of? Made of wood. This means that the set of antiderivatives of the integrand of the function “to be a door”, that is, its indefinite integral, is the function “to be a tree + C”, where C is a constant, which in this context can denote, for example, the type of tree. Just as a door is made from wood using some tools, a derivative of a function is “made” from an antiderivative function using formulas we learned while studying the derivative .

Then the table of functions of common objects and their corresponding antiderivatives (“to be a door” - “to be a tree”, “to be a spoon” - “to be metal”, etc.) is similar to the table of basic indefinite integrals, which will be given below. The table of indefinite integrals lists common functions with an indication of the antiderivatives from which these functions are “made”. In part of the problems on finding the indefinite integral, integrands are given that can be integrated directly without much effort, that is, using the table of indefinite integrals. In more complex problems, the integrand must first be transformed so that table integrals can be used.

Fact 2. When restoring a function as an antiderivative, we must take into account an arbitrary constant (constant) C, and in order not to write a list of antiderivatives with various constants from 1 to infinity, you need to write a set of antiderivatives with an arbitrary constant C, for example, like this: 5 x³+C. So, an arbitrary constant (constant) is included in the expression of the antiderivative, since the antiderivative can be a function, for example, 5 x³+4 or 5 x³+3 and when differentiated, 4 or 3, or any other constant goes to zero.

Let us pose the integration problem: for this function f(x) find such a function F(x), whose derivative equal to f(x).

Example 1. Find the set of antiderivatives of a function

Solution. For this function, the antiderivative is the function

Function F(x) is called an antiderivative for the function f(x), if the derivative F(x) is equal to f(x), or, which is the same thing, differential F(x) is equal f(x) dx, i.e.

(2)

Therefore, the function is an antiderivative of the function. However, it is not the only antiderivative for . They also serve as functions

Where WITH– arbitrary constant. This can be verified by differentiation.

Thus, if there is one antiderivative for a function, then for it there is an infinite number of antiderivatives that differ by a constant term. All antiderivatives for a function are written in the above form. This follows from the following theorem.

Theorem (formal statement of fact 2). If F(x) – antiderivative for the function f(x) on some interval X, then any other antiderivative for f(x) on the same interval can be represented in the form F(x) + C, Where WITH– arbitrary constant.

In the next example, we turn to the table of integrals, which will be given in paragraph 3, after the properties of the indefinite integral. We do this before reading the entire table so that the essence of the above is clear. And after the table and properties, we will use them in their entirety during integration.

Example 2. Find sets of antiderivative functions:

Solution. We find sets of antiderivative functions from which these functions are “made”. When mentioning formulas from the table of integrals, for now just accept that there are such formulas there, and we will study the table of indefinite integrals itself a little further.

1) Applying formula (7) from the table of integrals for n= 3, we get

2) Using formula (10) from the table of integrals for n= 1/3, we have

3) Since

then according to formula (7) with n= -1/4 we find

It is not the function itself that is written under the integral sign f, and its product by the differential dx. This is done primarily in order to indicate by which variable the antiderivative is sought. For example,

, ;

here in both cases the integrand is equal to , but its indefinite integrals in the cases considered turn out to be different. In the first case, this function is considered as a function of the variable x, and in the second - as a function of z .

The process of finding the indefinite integral of a function is called integrating that function.

Geometric meaning of the indefinite integral

Suppose we need to find a curve y=F(x) and we already know that the tangent of the tangent angle at each of its points is a given function f(x) abscissa of this point.

According to the geometric meaning of the derivative, the tangent of the tangent angle at a given point of the curve y=F(x) equal to the value of the derivative F"(x). So we need to find such a function F(x), for which F"(x)=f(x). Function required in the task F(x) is an antiderivative of f(x). The conditions of the problem are satisfied not by one curve, but by a family of curves. y=F(x)- one of such curves, and any other curve can be obtained from it by parallel translation along the axis Oy.

Let's call the graph of the antiderivative function of f(x) integral curve. If F"(x)=f(x), then the graph of the function y=F(x) there is an integral curve.

Fact 3. The indefinite integral is geometrically represented by the family of all integral curves , as in the picture below. The distance of each curve from the origin of coordinates is determined by an arbitrary integration constant C.

Properties of the indefinite integral

Fact 4. Theorem 1. The derivative of an indefinite integral is equal to the integrand, and its differential is equal to the integrand.

Fact 5. Theorem 2. Indefinite integral of the differential of a function f(x) is equal to the function f(x) up to a constant term , i.e.

(3)

Theorems 1 and 2 show that differentiation and integration are mutually inverse operations.

Fact 6. Theorem 3. The constant factor in the integrand can be taken out of the sign of the indefinite integral , i.e.

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

Table of antiderivatives ("integrals"). Tabular indefinite integrals. (The simplest integrals and integrals with a parameter).
Integral of a power function.	Integral of a power function.
An integral that reduces to the integral of a power function if x is driven under the differential sign.
	Integral of an exponential, where a is a constant number.
Integral of a complex exponential function.	Integral of an exponential function.
An integral equal to the natural logarithm.	Integral: "Long logarithm".
	Integral: "Long logarithm".
Integral: "High logarithm".	An integral, where x in the numerator is placed under the differential sign (the constant under the sign can be either added or subtracted), is ultimately similar to an integral equal to the natural logarithm.
Integral: "High logarithm".
Cosine integral.	Sine integral.
Integral equal to tangent.	Integral equal to cotangent.
Integral equal to both arcsine and arccosine
An integral equal to both arcsine and arccosine.	An integral equal to both arctangent and arccotangent.
Integral equal to cosecant.	Integral equal to secant.
Integral equal to arcsecant.	Integral equal to arccosecant.
Integral equal to arcsecant.	Integral equal to arcsecant.
Integral equal to the hyperbolic sine.	Integral equal to hyperbolic cosine.
Integral equal to the hyperbolic sine, where sinhx is the hyperbolic sine in the English version.	Integral equal to the hyperbolic cosine, where sinhx is the hyperbolic sine in the English version.
Integral equal to the hyperbolic tangent.	Integral equal to the hyperbolic cotangent.
Integral equal to the hyperbolic secant.	Integral equal to the hyperbolic cosecant.

Formulas for integration by parts. Integration rules.

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

Table of derivatives. Tabular derivatives. Derivative of the product. Derivative of the quotient. Derivative of a complex function.

If x is an independent variable, then:

Table of derivatives. Tabular derivatives."table derivative" - ​​yes, unfortunately, this is exactly how they are searched for on the Internet
	Derivative of a power function
	Derivative of the exponent
Derivative of a complex exponential function	Derivative of exponential function
Derivative of a logarithmic function	Derivative of the natural logarithm
	Derivative of the natural logarithm of a function
Derivative of sine	Derivative of cosine
Derivative of cosecant	Derivative of a secant
Derivative of arcsine	Derivative of arc cosine
Derivative of arcsine	Derivative of arc cosine
Tangent derivative	Derivative of cotangent
Derivative of arctangent	Derivative of arc cotangent
Derivative of arctangent	Derivative of arc cotangent
Derivative of arcsecant	Derivative of arccosecant
Derivative of arcsecant	Derivative of arccosecant
Derivative of the hyperbolic sine Derivative of the hyperbolic sine in the English version	Derivative of hyperbolic cosine Derivative of hyperbolic cosine in English version
Derivative of hyperbolic tangent	Derivative of hyperbolic cotangent
Derivative of the hyperbolic secant	Derivative of the hyperbolic cosecant

Rules of differentiation. Derivative of the product. Derivative of the quotient. Derivative of a complex function.
Derivative of a product (function) by a constant:
Derivative of sum (functions):
Derivative of the product (functions):
Derivative of the quotient (of functions):
Derivative of a complex function:

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

Basic logarithmic identity
Let's 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 to base e = 2.718281828459045...) ln(e)=1; ln(1)=0

Taylor series. Taylor series expansion of a function.

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

When using series called Taylor's rows mixed functions containing, say, algebraic, trigonometric and exponential functions can be expressed as purely algebraic functions. Using series, you can often quickly perform differentiation and integration.

The Taylor series in the neighborhood of point a has the form:

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)

The 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 (=McLaren) series (the expansion occurs around the point a=0)

at a=0

members of the series are determined by the formula

Conditions for using Taylor series.

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

2. It is necessary that derivatives exist for a given function at the point in the vicinity of which we are going to construct the Taylor series.

Properties of Taylor series.

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

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

Taylor series are used in approximation (approximation is a scientific method that consists of replacing some objects with others, in one sense or another close to the original ones, but simpler) of a function 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 the analysis of a linear system, in some sense equivalent to the original one.) equations occurs by expanding into a Taylor series and cutting off all 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 McLaren series.

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

Examples of some common Taylor series expansions in the vicinity of point 1

Principal integrals that every student should know

The listed integrals are the basis, the basis of the fundamentals. These formulas should definitely 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). Don't forget to add an arbitrary constant C to your answer when integrating!

Integral of a constant

∫ A d x = A x + C (1)

Integrating a Power Function

In fact, it was possible to limit ourselves to only formulas (5) and (7), but the rest of the integrals from this group occur so often 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 = ln | 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 exponential functions and hyperbolic functions

Of course, formula (8) (perhaps the most convenient for memorization) 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 simply remember these relations.

∫ 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 is that 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 function sinx is equal to cosx. This is not true! The integral of sine is equal to “minus cosine”, but the integral of cosx is equal to “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 that reduce to inverse trigonometric functions

Formula (16), leading to the arctangent, 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

It is also advisable to remember these formulas. They are also used quite often, and their output is quite tedious.

∫ 1 x 2 + a 2 d x = ln | x + x 2 + a 2 | +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 + x 2 + a 2 | + 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 rules of integration

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 an antiderivative for the function f(x). Please note: 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 product cannot be integrated. It’s just that every time you see an integral like (30), you will have to invent a way to “fight” it. In some cases, integration by parts will help you, in others you will have to make a change of variable, and sometimes even “school” algebra or trigonometry formulas can help.

A simple example of calculating the indefinite integral

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

Let us 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 obtain: ∫ 3 x 2 d x + ∫ 2 sin x d x − ∫ 7 e x d x + ∫ 12 d x

Let us remember 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, exponential 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 by 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 = ln | 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 + x 2 + a 2 | +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 + x 2 + a 2 | + 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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