Table of standard integrals and basic integration techniques. Basic formulas and methods of integration

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 angle of inclination of the tangent 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 these 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.

Basic formulas and methods of integration. The rule for integrating a sum or difference. Moving the constant outside the integral sign. Variable replacement method. Formula for integration by parts. An example of solving a problem.

The four main methods of integration are listed below.

1) The rule for integrating a sum or difference.
.
Here and below u, v, w are functions of the integration variable x.

2) Moving the constant outside the integral sign.
Let c be a constant independent of x.

3) Then it can be taken out of the integral sign.
Variable replacement method.
Let's consider the indefinite integral. If we can find such a function φ(x)
,
from x, so
.

4) then, by replacing the variable t = φ(x) , we have
,
Formula for integration by parts.

where u and v are functions of the integration variable.
The ultimate goal of calculating indefinite integrals is, through transformations, to reduce a given integral to the simplest integrals, which are called tabular integrals. Table integrals are expressed through elementary functions using known formulas.

See Table of Integrals >>>

Example

Calculate indefinite integral

Solution
We note that the integrand is the sum and difference of three terms:
, And . 1 .

Applying the method 5, 4, Next, we note that the integrands of the new integrals are multiplied by constants 2 And 2 .

, respectively. Applying the method
.
In the table of integrals we find the formula 2 Assuming n =

, we find the first integral.
.
Let us rewrite the second integral in the form

We notice that . Then.
.
Let's use the third method. We change the variable t = φ

(x) = ln x

In the table of integrals we find the formula
.
Since the variable of integration can be denoted by any letter, then
Let us rewrite the third integral in the form
We apply the integration by parts formula.
;
;

;
;
.

Let's put it.

Then

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:).

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 out the constant factor 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.

Let's look at a few more examples.

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 sum cube formula , and then divide the resulting polynomial by the denominator, term by term.

=((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

On this page you will find:

1. Actually, the table of antiderivatives - it can be downloaded in PDF format and printed;

2. Video on how to use this table;

3. A bunch of examples of calculating the antiderivative from various textbooks and tests.

In the video itself, we will analyze many problems where you need to calculate antiderivatives of functions, often quite complex, but most importantly, they are not power functions. All functions summarized in the table proposed above must be known by heart, like derivatives. Without them, further study of integrals and their application to solve practical problems is impossible.

Today we continue to study primitives and move on to a slightly more complex topic. If last time we considered antiderivatives only of power functions and slightly more complex constructions, today we will look at trigonometry and much more.

As I said in the last lesson, antiderivatives, unlike derivatives, are never solved “outright” using any standard rules. Moreover, the bad news is that, unlike the derivative, the antiderivative may not be considered at all. If we write a completely random function and try to find its derivative, then with a very high probability we will succeed, but the antiderivative will almost never be calculated in this case. But there is good news: there is a fairly large class of functions called elementary functions, the antiderivatives of which are very easy to calculate. And all the other more complex structures that are given on all kinds of tests, independent tests and exams, in fact, are made up of these elementary functions through addition, subtraction and other simple actions. The prototypes of such functions have long been calculated and compiled into special tables. It is these functions and tables that we will work with today.

But we will begin, as always, with a repetition: let’s remember what an antiderivative is, why there are infinitely many of them, and how to determine their general appearance. To do this, I picked up two simple problems.

Solving easy examples

Example #1

Let us immediately note that $\frac(\text( )\!\!\pi\!\!\text( ))(6)$ and in general the presence of $\text( )\!\!\pi\!\!\ text( )$ immediately hints to us that the required antiderivative of the function is related to trigonometry. And, indeed, if we look at the table, we will find that $\frac(1)(1+((x)^(2)))$ is nothing more than $\text(arctg)x$. So let's write it down:

In order to find, you need to write down the following:

\[\frac(\pi )(6)=\text(arctg)\sqrt(3)+C\]

\[\frac(\text( )\!\!\pi\!\!\text( ))(6)=\frac(\text( )\!\!\pi\!\!\text( )) (3)+C\]

Example No. 2

We are also talking about trigonometric functions here. If we look at the table, then, indeed, this is what happens:

We need to find, among the entire set of antiderivatives, the one that passes through the indicated point:

\[\text( )\!\!\pi\!\!\text( )=\arcsin \frac(1)(2)+C\]

\[\text( )\!\!\pi\!\!\text( )=\frac(\text( )\!\!\pi\!\!\text( ))(6)+C\]

Let's finally write it down:

It's that simple. The only problem is that in order to calculate antiderivatives of simple functions, you need to learn a table of antiderivatives. However, after studying the derivative table for you, I think this will not be a problem.

Solving problems containing an exponential function

To begin with, let's write the following formulas:

\[((e)^(x))\to ((e)^(x))\]

\[((a)^(x))\to \frac(((a)^(x)))(\ln a)\]

Let's see how this all works in practice.

Example #1

If we look at the contents of the brackets, we will notice that in the table of antiderivatives there is no such expression for $((e)^(x))$ to be in a square, so this square must be expanded. To do this, we use the abbreviated multiplication formulas:

Let's find the antiderivative for each of the terms:

\[((e)^(2x))=((\left(((e)^(2)) \right))^(x))\to \frac(((\left(((e)^ (2)) \right))^(x)))(\ln ((e)^(2)))=\frac(((e)^(2x)))(2)\]

\[((e)^(-2x))=((\left(((e)^(-2)) \right))^(x))\to \frac(((\left(((e )^(-2)) \right))^(x)))(\ln ((e)^(-2)))=\frac(1)(-2((e)^(2x))) \]

Now let’s collect all the terms into a single expression and get the general antiderivative:

Example No. 2

This time the degree is larger, so the abbreviated multiplication formula will be quite complex. So let's open the brackets:

Now let’s try to take the antiderivative of our formula from this construction:

As you can see, there is nothing complicated or supernatural in the antiderivatives of the exponential function. All of them are calculated through tables, but attentive students will probably notice that the antiderivative $((e)^(2x))$ is much closer to simply $((e)^(x))$ than to $((a)^(x ))$. So, maybe there is some more special rule that allows, knowing the antiderivative $((e)^(x))$, to find $((e)^(2x))$? Yes, such a rule exists. And, moreover, it is an integral part of working with the table of antiderivatives. We will now analyze it using the same expressions that we just worked with as an example.

Rules for working with the table of antiderivatives

Let's write our function again:

In the previous case, we used the following formula to solve:

\[((a)^(x))\to \frac(((a)^(x)))(\operatorname(lna))\]

But now let’s do it a little differently: let’s remember on what basis $((e)^(x))\to ((e)^(x))$. As I already said, because the derivative $((e)^(x))$ is nothing more than $((e)^(x))$, therefore its antiderivative will be equal to the same $((e) ^(x))$. But the problem is that we have $((e)^(2x))$ and $((e)^(-2x))$. Now let's try to find the derivative of $((e)^(2x))$:

\[((\left(((e)^(2x)) \right))^(\prime ))=((e)^(2x))\cdot ((\left(2x \right))^( \prime ))=2\cdot ((e)^(2x))\]

Let's rewrite our construction again:

\[((\left(((e)^(2x)) \right))^(\prime ))=2\cdot ((e)^(2x))\]

\[((e)^(2x))=((\left(\frac(((e)^(2x)))(2) \right))^(\prime ))\]

This means that when we find the antiderivative $((e)^(2x))$ we get the following:

\[((e)^(2x))\to \frac(((e)^(2x)))(2)\]

As you can see, we got the same result as before, but we did not use the formula to find $((a)^(x))$. Now this may seem stupid: why complicate the calculations when there is a standard formula? However, in slightly more complex expressions you will find that this technique is very effective, i.e. using derivatives to find antiderivatives.

As a warm-up, let's find the antiderivative of $((e)^(2x))$ in a similar way:

\[((\left(((e)^(-2x)) \right))^(\prime ))=((e)^(-2x))\cdot \left(-2 \right)\]

\[((e)^(-2x))=((\left(\frac(((e)^(-2x)))(-2) \right))^(\prime ))\]

When calculating, our construction will be written as follows:

\[((e)^(-2x))\to -\frac(((e)^(-2x)))(2)\]

\[((e)^(-2x))\to -\frac(1)(2\cdot ((e)^(2x)))\]

We got exactly the same result, but took a different path. It is this path, which now seems a little more complicated to us, that in the future will turn out to be more effective for calculating more complex antiderivatives and using tables.

Note! This is a very important point: antiderivatives, like derivatives, can be counted in many different ways. However, if all calculations and calculations are equal, then the answer will be the same. We have just seen this in the example of $((e)^(-2x))$ - on the one hand, we calculated this antiderivative “right through”, using the definition and calculating it using transformations, on the other hand, we remembered that $ ((e)^(-2x))$ can be represented as $((\left(((e)^(-2)) \right))^(x))$ and only then we used the antiderivative for the function $( (a)^(x))$. However, after all the transformations, the result was the same, as expected.

And now that we understand all this, it’s time to move on to something more significant. Now we will analyze two simple constructions, but the technique that will be used when solving them is a more powerful and useful tool than simply “running” between neighboring antiderivatives from the table.

Problem solving: finding the antiderivative of a function

Example #1

Let's break down the amount that is in the numerators into three separate fractions:

This is a fairly natural and understandable transition - most students do not have problems with it. Let's rewrite our expression as follows:

Now let's remember this formula:

In our case we will get the following:

To get rid of all these three-story fractions, I suggest doing the following:

Example No. 2

Unlike the previous fraction, the denominator is not a product, but a sum. In this case, we can no longer divide our fraction into the sum of several simple fractions, but we must somehow try to make sure that the numerator contains approximately the same expression as the denominator. In this case, it's quite simple to do it:

This notation, which in mathematical language is called “adding a zero,” will allow us to again divide the fraction into two pieces:

Now let's find what we were looking for:

That's all the calculations. Despite the apparent greater complexity than in the previous problem, the amount of calculations turned out to be even smaller.

Nuances of the solution

And this is where the main difficulty of working with tabular antiderivatives lies, this is especially noticeable in the second task. The fact is that in order to select some elements that are easily calculated through the table, we need to know what exactly we are looking for, and it is in the search for these elements that the entire calculation of antiderivatives consists.

In other words, it is not enough just to memorize the table of antiderivatives - you need to be able to see something that does not yet exist, but what the author and compiler of this problem meant. That is why many mathematicians, teachers and professors constantly argue: “What is taking antiderivatives or integration - is it just a tool or is it a real art?” In fact, in my personal opinion, integration is not an art at all - there is nothing sublime in it, it is just practice and more practice. And to practice, let's solve three more serious examples.

We train in integration in practice

Task No. 1

Let's write the following formulas:

\[((x)^(n))\to \frac(((x)^(n+1)))(n+1)\]

\[\frac(1)(x)\to \ln x\]

\[\frac(1)(1+((x)^(2)))\to \text(arctg)x\]

Let's write the following:

Problem No. 2

Let's rewrite it as follows:

The total antiderivative will be equal to:

Task No. 3

The difficulty of this task is that, unlike the previous functions above, there is no variable $x$ at all, i.e. it is not clear to us what to add or subtract in order to get at least something similar to what is below. However, in fact, this expression is considered even simpler than any of the previous expressions, because this function can be rewritten as follows:

You may now ask: why are these functions equal? Let's check:

Let's rewrite it again:

Let's transform our expression a little:

And when I explain all this to my students, almost always the same problem arises: with the first function everything is more or less clear, with the second you can also figure it out with luck or practice, but what kind of alternative consciousness do you need to have in order to solve the third example? Actually, don't be scared. The technique that we used when calculating the last antiderivative is called “decomposition of a function into its simplest”, and this is a very serious technique, and a separate video lesson will be devoted to it.

In the meantime, I propose to return to what we just studied, namely, to exponential functions and somewhat complicate the problems with their content.

More complex problems for solving antiderivative exponential functions

Task No. 1

Let's note the following:

\[((2)^(x))\cdot ((5)^(x))=((\left(2\cdot 5 \right))^(x))=((10)^(x) )\]

To find the antiderivative of this expression, simply use the standard formula - $((a)^(x))\to \frac(((a)^(x)))(\ln a)$.

In our case, the antiderivative will be like this:

Of course, compared to the design we just solved, this one looks simpler.

Problem No. 2

Again, it's easy to see that this function can easily be divided into two separate terms - two separate fractions. Let's rewrite:

It remains to find the antiderivative of each of these terms using the formula described above:

Despite the apparent greater complexity of exponential functions compared to power functions, the overall volume of calculations and calculations turned out to be much simpler.

Of course, for knowledgeable students, what we have just discussed (especially against the backdrop of what we have analyzed before) may seem like elementary expressions. However, when choosing these two problems for today's video lesson, I did not set myself the goal of telling you another complex and sophisticated technique - all I wanted to show you is that you should not be afraid to use standard algebra techniques to transform original functions.

Using a "secret" technique

In conclusion, I would like to look at another interesting technique, which, on the one hand, goes beyond the scope of what we mainly discussed today, but, on the other hand, it is, firstly, not at all complicated, i.e. Even beginner students can master it, and, secondly, it is quite often found in all kinds of tests and independent work, i.e. knowledge of it will be very useful in addition to knowledge of the table of antiderivatives.

Task No. 1

Obviously, we have something very similar to a power function. What should we do in this case? Let's think about it: $x-5$ is not that much different from $x$ - they just added $-5$. Let's write it like this:

\[((x)^(4))\to \frac(((x)^(5)))(5)\]

\[((\left(\frac(((x)^(5)))(5) \right))^(\prime ))=\frac(5\cdot ((x)^(4))) (5)=((x)^(4))\]

Let's try to find the derivative of $((\left(x-5 \right))^(5))$:

\[((\left(((\left(x-5 \right))^(5)) \right))^(\prime ))=5\cdot ((\left(x-5 \right)) ^(4))\cdot ((\left(x-5 \right))^(\prime ))=5\cdot ((\left(x-5 \right))^(4))\]

This implies:

\[((\left(x-5 \right))^(4))=((\left(\frac(((\left(x-5 \right))^(5)))(5) \ right))^(\prime ))\]

There is no such value in the table, so we have now derived this formula ourselves using the standard antiderivative formula for a power function. Let's write the answer like this:

Problem No. 2

Many students who look at the first solution may think that everything is very simple: just replace $x$ in the power function with a linear expression, and everything will fall into place. Unfortunately, everything is not so simple, and now we will see this.

By analogy with the first expression, we write the following:

\[((x)^(9))\to \frac(((x)^(10)))(10)\]

\[((\left(((\left(4-3x \right))^(10)) \right))^(\prime ))=10\cdot ((\left(4-3x \right)) ^(9))\cdot ((\left(4-3x \right))^(\prime ))=\]

\[=10\cdot ((\left(4-3x \right))^(9))\cdot \left(-3 \right)=-30\cdot ((\left(4-3x \right)) ^(9))\]

Returning to our derivative, we can write:

\[((\left(((\left(4-3x \right))^(10)) \right))^(\prime ))=-30\cdot ((\left(4-3x \right) )^(9))\]

\[((\left(4-3x \right))^(9))=((\left(\frac(((\left(4-3x \right))^(10)))(-30) \right))^(\prime ))\]

This immediately follows:

Nuances of the solution

Please note: if nothing essentially changed last time, then in the second case, instead of $-10$, $-30$ appeared. What is the difference between $-10$ and $-30$? Obviously, by a factor of $-3$. Question: where did it come from? If you look closely, you can see that it was taken as a result of calculating the derivative of a complex function - the coefficient that stood at $x$ appears in the antiderivative below. This is a very important rule, which I initially did not plan to discuss at all in today’s video lesson, but without it the presentation of tabular antiderivatives would be incomplete.

So let's do it again. Let there be our main power function:

\[((x)^(n))\to \frac(((x)^(n+1)))(n+1)\]

Now, instead of $x$, let's substitute the expression $kx+b$. What will happen then? We need to find the following:

\[((\left(kx+b \right))^(n))\to \frac(((\left(kx+b \right))^(n+1)))(\left(n+ 1\right)\cdot k)\]

On what basis do we claim this? Very simple. Let's find the derivative of the construction written above:

\[((\left(\frac(((\left(kx+b \right))^(n+1)))(\left(n+1 \right)\cdot k) \right))^( \prime ))=\frac(1)(\left(n+1 \right)\cdot k)\cdot \left(n+1 \right)\cdot ((\left(kx+b \right))^ (n))\cdot k=((\left(kx+b \right))^(n))\]

This is the same expression that originally existed. Thus, this formula is also correct, and it can be used to supplement the table of antiderivatives, or it is better to simply memorize the entire table.

Conclusions from the “secret: technique:

• Both functions that we just looked at can, in fact, be reduced to the antiderivatives indicated in the table by expanding the degrees, but if we can more or less somehow cope with the fourth degree, then I wouldn’t even consider the ninth degree dared to reveal.
• If we were to expand the degrees, we would end up with such a volume of calculations that a simple task would take us an inappropriately large amount of time.
• That is why such problems, which contain linear expressions, do not need to be solved “headlong”. As soon as you come across an antiderivative that differs from the one in the table only by the presence of the expression $kx+b$ inside, immediately remember the formula written above, substitute it into your table antiderivative, and everything will turn out much faster and easier.

Naturally, due to the complexity and seriousness of this technique, we will return to its consideration many times in future video lessons, but that’s all for today. I hope this lesson will really help those students who want to understand antiderivatives and integration.

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

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

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)

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

∫ 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 |