Peelback and its properties
An antiderivative of a function f(x) on the interval (a; b) is a function F(x) such that the equality holds for any x from the given interval.
If we take into account the fact that the derivative of the constant C is equal to zero, then the equality is true 
. Thus, the function f(x) has a set of antiderivatives F(x)+C, for an arbitrary constant C, and these antiderivatives differ from each other by an arbitrary constant value.
Properties of the antiderivative.
If the function F(x) is an antiderivative for the function f(x) on the interval X, then the function f(x) + C, where C is an arbitrary constant, will also be an antiderivative for f(x) on this interval.
If the function F(x) is some antiderivative of the function f(x) on the interval X=(a,b), then any other antiderivative F1(x) can be represented as F1(x) = F(x) + C, where C is a constant function on X.
2 Definition of indefinite integral.
The entire set of antiderivatives of the function f(x) is called the indefinite integral of this function and is denoted 
.
The expression is called the integrand, and f(x) is called the integrand. The integrand represents the differential of the function f(x).
The action of finding an unknown function given its differential is called indefinite integration, because the result of integration is not one function F(x), but a set of its antiderivatives F(x)+C.
properties of the indefinite integral (properties of the antiderivative).
The derivative of the integration result is equal to the integrand.
The indefinite integral of the differential of a function is equal to the sum of the function itself and an arbitrary constant.
where k is an arbitrary constant. The coefficient can be taken out as a sign of the indefinite integral.
The indefinite integral of the sum/difference of functions is equal to the sum/difference of the indefinite integrals of functions.
Changing a Variable in an Indefinite Integral
Variable replacement in the indefinite integral is performed using two types of substitutions:
a) where is a monotone, continuously differentiable function of the new variable t. The variable replacement formula in this case is:
Where U is a new variable. The formula for replacing a variable with this substitution is:
Integration by parts
Finding the integral using the formula 
is called integration by parts. Here U=U(x),υ=υ(x) are continuously differentiable functions of x. Using this formula, finding an integral is reduced to finding another integral; its use is advisable in cases where the last integral is either simpler than the original one or similar to it.
In this case, υ is taken to be a function that simplifies upon differentiation, and dU is taken to be that part of the integrand whose integral is known or can be found.
Newton–Leibniz formula
Continuity of the definite integral as a function of the upper limit
If the function y = f (x) is integrable on the interval , then, obviously, it is also integrable on an arbitrary interval [a, x] embedded in . Function 
,
where x О is called an integral with a variable upper limit. The value of the function Ф (x) at point x is equal to the area S(x) under the curve y = f (x) on the segment [a, x]. This is the geometric meaning of an integral with a variable upper limit.
Theorem. If the function f (x) is continuous on an interval, then the function Ф (x) is also continuous on [a, b].
Let Δx be such that x + Δ x О . We have
By the mean value theorem, there is a value with О [ x, x + Δ x] such that Since with О , and the function f (x) is bounded, then passing to the limit as Δ x → 0, we obtain
ODR 1st order
What is the difference between homogeneous differential equations and other types of differential equations? The easiest way to immediately explain this is with a specific example.
Solve differential equation
What should you analyze first when solving any first order differential equation? First of all, it is necessary to check whether it is possible to immediately separate the variables using “school” actions? Usually this analysis is done mentally or by trying to separate the variables in a draft.
In this example, the variables cannot be divided (you can try throwing terms from part to part, raising factors out of brackets, etc.). By the way, in this example, the fact that the variables cannot be divided is quite obvious due to the presence of the multiplier
The question arises: how to solve this diffuse problem?
It is necessary to check whether this equation is homogeneous? The verification is simple, and the verification algorithm itself can be formulated as follows:
To the original equation:
Instead of x we substitute instead of y we substitute the derivative without touching: 
The letter lambda is some abstract numerical parameter, the point is not in the lambdas themselves, and not in their values, but the point is this:
If, as a result of transformations, it is possible to reduce ALL “lambdas” (i.e., obtain the original equation), then this differential equation is homogeneous.
It is obvious that lambdas are immediately reduced by the exponent: 
Now on the right side we take the lambda out of brackets: 
Both sides of the equation can be reduced to this same lambda: As a result, all the lambdas disappeared like a dream, like morning fog, and we got the original equation.
Conclusion: This equation is homogeneous
LOU.General properties of solutions
that is, it is linear with respect to the unknown function y and its derivatives and . The coefficients and and the right side of this equation are continuous.
If the right side of the equation is , then the equation is called linear inhomogeneous. If , then the equation has the form
 | (9) |
and is called linear homogeneous.
Let and be some particular solutions of equation (9), that is, do not contain arbitrary constants.
Theorem 1. If and are two partial solutions of a linear homogeneous equation of the second order, then they are also a solution to this equation.
Since and are solutions to equation (9), they turn this equation into an identity, that is
 And  | (10) |
Let's substitute into equation (9). Then we have:
By virtue of (10). This means that this is a solution to the equation.
Theorem 2. If is a solution to a linear homogeneous equation of the second order, and C–constant, then is also a solution to this equation.
Proof. Let's substitute into equation (9). We get: that is, a solution to the equation.
Consequence. If and are solutions to equation (9), then they are also its solutions by virtue of Theorems (1) and (2).
Definition. Two solutions and equations (9) are called linearly dependent (on the interval) if it is possible to select numbers and that are not simultaneously equal to zero such that the linear combination of these solutions is identically equal to zero on , that is, if .
If such numbers cannot be selected, then the solutions are called linearly independent (on the segment ).
Obviously, solutions and will be linearly dependent if and only if their ratio is constant, that is (or vice versa).
In fact, if and are linearly dependent, then where is at least one constant or nonzero. Let, for example, . Then , , Denoting we get , that is, the relationship is constant.
Back if then 
. Here the coefficient at , that is, is different from zero, which by definition means that and are linearly dependent.
Comment. From the definition of linearly independent solutions and the reasoning above, we can conclude that if and are linearly independent, then their ratio cannot be constant.
For example, the functions and at are linearly independent, since 
, because . Here are the 5 functions x And x– are linearly dependent, since their ratio is .
Theorem. If and are linearly independent partial solutions of a linear homogeneous equation of the second order, then their linear combination , where and are arbitrary constants, is a general solution to this equation.
Proof. By virtue of Theorems 1 and 2 (and their corollaries), it is a solution to equation (9) for any choice of constants and .
If the solutions and are linearly independent, then – is a general solution, since this solution contains two arbitrary constants that cannot be reduced to one.
At the same time, even if they were linearly dependent solutions, it would no longer be a general solution. In this case, where α -constant. Then , where is a constant. cannot be a general solution to a second-order differential equation, since it depends on only one constant.
So, the general solution to equation (9):
19. The concept of a linearly independent system of functions. Vronsky's determinant. sufficient condition for linear independence. concept of a fundamental system of function. Examples. Necessary and sufficient condition for the Wronski determinant to differ from zero on the interval [a,c]
The concept of a linearly independent system of functions 
Functions 
are called linearly dependent on if one of them is a linear combination of the others. In other words, the functions are called linearly dependent on if there are numbers of which at least one is not equal to zero such that
If identity (4) is satisfied only in the case when all , then the functions are called linearly independent on .
System of linearly independent solutions on an interval
homogeneous differential equation of the th order (3) with continuous coefficients is called the fundamental system of solutions to this equation.
To solve a linear homogeneous differential equation of the th order (3) with continuous coefficients, it is necessary to find its fundamental system of solutions.
According to Theorem 1, an arbitrary linear combination of solutions, i.e., the sum
, (5)
where are arbitrary numbers, is, in turn, a solution to equation (3) on . But it turns out that, conversely, every solution of the differential equation (3) on the interval is some linear combination of the indicated (independent) partial solutions (see Theorem 4 below), forming a fundamental system of solutions.
Thus, the general solution of the homogeneous differential equation (3) has the form (5), where are arbitrary constants, and are partial solutions (3), forming a fundamental system of solutions to the homogeneous equation.
Note that the general solution of the inhomogeneous equation (1) is the sum of any particular solution of it and the general solution of the homogeneous equation
. (6)
Indeed,
.
On the other hand, if there is an arbitrary solution to equation (1), then
and, therefore, is a solution to a homogeneous equation; but then there are numbers such that
,
i.e., equality (6) holds for these numbers.
Vronsky's determinant.
Theorem 2. If the functions are linearly dependent on and have derivatives up to the th order, then the determinant
. (7)
I
The determinant (7) is called the Wronski determinant or Wronskian and is denoted by the symbol 
.
Proof. Since the functions are linearly dependent on , then there exist numbers that are not all equal to zero for which identity (4) holds on . Differentiating it once, we obtain a system of equations
By condition, this homogeneous system has a nontrivial solution (i.e., at least one) for . The latter is possible when the determinant of the system, which is the Wronski determinant, is identically equal to zero. The theorem has been proven.
Comment. It follows from Theorem 2 that if at at least one point , then the functions are linearly independent on .
Example 2. The functions are linearly independent on any , since
.
Example 3. The functions are linearly independent on any if - different numbers (real or complex).
Indeed.
,
since the last determinant is the Vandermonde determinant, which is not equal to zero for different ones.
Example 4: Functions 
are linearly independent at any .
Since
then the linear independence of these functions follows from the second example.
Theorem 3. In order for solutions 
linear differential homogeneous equation with continuous coefficients are linearly independent on , it is necessary and sufficient that for all .
Proof. 1) If on , then the functions are linearly independent regardless of whether they are solutions to the equation or not (see remark).
2) Let be linearly independent functions on and be solutions of the equation .
Let us prove that everywhere on . Let us assume the opposite, that there is a point at which . Let us choose numbers that are not equal to zero at the same time, so that they are solutions of the system
(8)
This can be done since the determinant of system (8) is . Then, by Theorem 1, the function 
will be a solution to the equation with zero initial conditions (according to (8))
But the trivial solution also satisfies the same conditions. By virtue of the existence and uniqueness theorem, there can be only one solution that satisfies these initial conditions, therefore, on i.e., the functions are linearly dependent on , which was not assumed. The theorem has been proven.
If are discontinuous functions in the interval where we are looking for a solution, then the equation may have more than one solution that satisfies the initial conditions, and then it is possible that on .
Example 5. It's easy to check that the functions
are linearly independent on and for them on .
This is due to the fact that the function 
is a general solution to the equation
,
Where 
discontinuous at point . For this equation, the existence and uniqueness theorem does not hold (in the neighborhood of the point ). Not only the function, but also the function is a solution to a differential equation that satisfies the conditions and for .
Structure of the general solution.
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Theorem 4. If are linearly independent solutions of a linear homogeneous differential equation of the th order with continuous coefficients 
, then the function
, (9)
where are arbitrary constants, is a general solution of the equation , i.e., the sum (9) for any , is a solution to this equation and, conversely, any solution to this equation can be represented as a sum (9) for the corresponding values of .
Proof. We already know that the sum (9) for any is a solution to the equation . Let, conversely, there be an arbitrary solution to this equation. Let's put
For the numbers received 
let's compose a linear system of equations for unknown numbers: , it is enough to find some real constants. To find a general solution to equation (8), we do this. We compose the characteristic equation for equation (8): . Using the initial conditions, we determine
Consider the linear differential equation n-th order
y (n) + a n -1 (x)y (n- 1) + ... + a 1 (x)y" + a 0 (x)y = f(x).
with continuous coefficients a n -1 (x), a n -2 (x), ..., a 1 (x), a 0 (x) and continuous right-hand side f(x).
Superposition principle based on the following properties of solutions of linear differential equations.
1. If y 1 (x) And y 2 (x) - two solutions to a linear homogeneous differential equation
y (n) + a n -1 (x)y (n- 1) + ... + a 1 (x)y" + a 0 (x)y = 0
then any linear combination of them y(x) = C 1 y 1 (x) + C 2 y 2 (x) is a solution to this homogeneous equation.
2. If y 1 (x) And y 2 (x) - two solutions to a linear inhomogeneous equation L(y) = f(x) then their difference y(x) = y 1 (x) − y 2 (x) is a solution to the homogeneous equation L(y) = 0 .
3. Any solution to an inhomogeneous linear equation L(y) = f(x) is the sum of any fixed (particular) solution of an inhomogeneous equation and some solution of a homogeneous equation.
4. If y 1 (x) And y 2 (x) - solutions of linear inhomogeneous equations L(y) = f 1 (x) And L(y) = f 2 (x) accordingly, then their sum y(x) =y 1 (x) + y 2 (x) is a solution to the inhomogeneous equation L(y) = f 1 (x) + f 2 (x).
Usually this last statement is called superposition principle.
Method of variation of constants
Consider the inhomogeneous equation of the th order
where the coefficients and the right side are given continuous functions on the interval.
Let us assume that we know the fundamental system of solutions corresponding homogeneous equation
As we showed in § 1.15 (formula (6)), the general solution to equation (1) is equal to the sum of the general solution to equation (2) and any solution to equation (1).
The solution to the inhomogeneous equation (1) can be
Change of variable in an indefinite integral is used to find integrals in which one of the functions is the derivative of another function. Let there be an integral $ \int f(x) dx $, let us make the replacement $ x=\phi(t) $. Note that the function $ \phi(t) $ is differentiable, so we can find $ dx = \phi"(t) dt $.
Now we substitute $ \begin(vmatrix) x = \phi(t) \\ dx = \phi"(t) dt \end(vmatrix) $ into the integral and we get that:
$$ \int f(x) dx = \int f(\phi(t)) \cdot \phi"(t) dt $$
This one is formula for changing a variable in an indefinite integral.
Algorithm of the variable replacement method
Thus, if the problem is given an integral of the form: $$ \int f(\phi(x)) \cdot \phi"(x) dx $$ It is advisable to replace the variable with a new one: $$ t = \phi(x) $ $ $$ dt = \phi"(t) dt $$
After this, the integral will be presented in a form that can be easily taken by the basic integration methods: $$ \int f(\phi(x)) \cdot \phi"(x) dx = \int f(t)dt $$
Don't forget to also return the replaced variable back to $x$.
Examples of solutions
| Example 1 |
|
Find the indefinite integral using the change of variable method: $$ \int e^(3x) dx $$ |
| Solution |
|
We replace the variable in the integral with $ t = 3x, dt = 3dx $: $$ \int e^(3x) dx = \int e^t \frac(dt)(3) = \frac(1)(3) \int e^t dt = $$ The integral of the exponential is still the same according to the integration table, although instead of $ x $ it is written $ t $: $$ = \frac(1)(3) e^t + C = \frac(1)(3) e^(3x) + C $$ If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner! |
| Answer |
| $$ \int e^(3x) dx = \frac(1)(3) e^(3x) + C $$ |
Integration by substitution (variable replacement). Suppose you need to calculate an integral that is not tabular. The essence of the substitution method is that in the integral the variable x is replaced by the variable t according to the formula x = q(t), from which dx = q"(t)dt.
Theorem. Let the function x=t(t) be defined and differentiable on a certain set T and let X be the set of values of this function on which the function f(x) is defined. Then if on the set X the function f(x) has an antiderivative, then on the set T the formula is valid:
Formula (1) is called the change of variable formula in the indefinite integral.
Integration by parts. The method of integration by parts follows from the formula for the differential of the product of two functions. Let u(x) and v(x) be two differentiable functions of the variable x. Then:
d(uv)=udv+vdu. - (3)
Integrating both sides of equality (3), we obtain:
But since then:
Relation (4) is called the integration by parts formula. Using this formula, find the integral. It is advisable to use it when the integral on the right side of formula (4) is simpler to calculate than the original one.
In formula (4) there is no arbitrary constant C, since on the right side of this formula there is an indefinite integral containing an arbitrary constant.
We present some frequently encountered types of integrals calculated by the method of integration by parts.
I. Integrals of the form, (P n (x) is a polynomial of degree n, k is a certain number). To find these integrals, it is enough to set u=P n (x) and apply formula (4) n times.
II. Integrals of the form, (Pn(x) is a polynomial of degree n with respect to x). They can be found using frequencies, taking for u a function that is a multiplier for P n (x).
Variable changes can be used to evaluate simple integrals and, in some cases, to simplify the calculation of more complex ones.
The variable replacement method is that we move from the original integration variable, let it be x, to another variable, which we denote as t. In this case, we believe that the variables x and t are related by some relation x = x (t), or t = t (x). For example, x = ln t, x = sin t, t = 2 x + 1, and so on. Our task is to select such a relationship between x and t that the original integral is either reduced to a tabular one or becomes simpler.
Basic variable replacement formula
Let's consider the expression that stands under the integral sign. It consists of the product of the integrand, which we denote as f (x) and differential dx: . Let us move to a new variable t by choosing some relation x = x (t). Then we must express the function f (x) and the differential dx through the variable t.
To express the integrand function f (x) through the variable t, you just need to substitute the selected relation x = x instead of the variable x (t).
The differential conversion is done like this:
.
That is, the differential dx is equal to the product of the derivative of x with respect to t and the differential dt.
Then
.
In practice, the most common case is in which we perform a replacement by choosing a new variable as a function of the old one: t = t (x). If we guessed that the integrand function can be represented as
,
where t′ (x) is the derivative of t with respect to x, then
.
So, the basic variable replacement formula can be presented in two forms.
(1)
,
where x is a function of t.
(2)
,
where t is a function of x.
Important Note
In tables of integrals, the integration variable is most often denoted as x. However, it is worth considering that the integration variable can be denoted by any letter. Moreover, any expression can be used as an integration variable.
As an example, consider the table integral
.
Here x can be replaced by any other variable or function of a variable. Here are examples of possible options:
;
;
.
In the last example, you need to take into account that when moving to the integration variable x, the differential is transformed as follows:
.
Then
.
This example captures the essence of integration by substitution. That is, we must guess that
.
After which the integral is reduced to a tabular one.
.
You can evaluate this integral using a change of variable using the formula (2)
. Let's put t = x 2+x. Then
;
;
.
Examples of integration by change of variable
1)
Let's calculate the integral
.
We notice that (sin x)′ = cos x. Then
.
Here we have used the substitution t = sin x.
2)
Let's calculate the integral
.
We notice that . Then
.
Here we performed the integration by changing the variable t = arctan x.
3)
Let's integrate
.
We notice that . Then
. Here, during integration, the variable t = x is replaced 2 + 1
.
Linear substitutions
Perhaps the most common are linear substitutions. This is a replacement for a variable of the form
t = ax + b,
where a and b are constants. With such a replacement, the differentials are related by the relation
.
Examples of integration by linear substitutions
A) Calculate integral
.
Solution.
.
B) Find the integral
.
Solution.
Let's use the properties of the exponential function.
.
ln 2- this is a constant. We calculate the integral.
.
C) Calculate integral
.
Solution.
Let us reduce the quadratic polynomial in the denominator of the fraction to the sum of squares.
.
We calculate the integral.
.
D) Find the integral
.
Solution.
Let's transform the polynomial under the root.
.
We integrate using the variable replacement method.
.
Previously we received the formula
.
From here
.
Substituting this expression, we get the final answer.
E) Calculate integral
.
Solution.
Let's apply the formula for the product of sine and cosine.
;
.
We integrate and make substitutions.
.
References:
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003.
In this lesson we will get acquainted with one of the most important and most common techniques that is used when solving indefinite integrals - the variable change method. Successful mastery of the material requires initial knowledge and integration skills. If there is a feeling of an empty full kettle in integral calculus, then you should first familiarize yourself with the material, where I explained in an accessible form what an integral is and analyzed in detail basic examples for beginners.
Technically, the method of changing a variable in an indefinite integral is implemented in two ways:
– Subsuming the function under the differential sign;
– Actually replacing the variable.
Essentially, these are the same thing, but the design of the solution looks different.
Let's start with a simpler case.
Subsuming a function under the differential sign
At the lesson Indefinite integral. Examples of solutions we learned how to open the differential, I remind you of the example I gave:
That is, revealing a differential is formally almost the same as finding a derivative.
Example 1
Perform check.
We look at the table of integrals and find a similar formula: 
. But the problem is that under the sine we have not just the letter “X”, but a complex expression. What to do?
We bring the function under the differential sign:
By opening the differential, it is easy to check that: 
In fact and 
is a recording of the same thing.
But, nevertheless, the question remained, how did we come to the idea that at the first step we need to write our integral exactly like this: 
? Why is this and not otherwise?
Formula 
(and all other table formulas) are valid and applicable NOT ONLY for the variable, but also for any complex expression ONLY AS A FUNCTION ARGUMENT( – in our example) AND THE EXPRESSION UNDER THE DIFFERENTIAL SIGN WERE THE SAME
.
Therefore, the mental reasoning when solving should be something like this: “I need to solve the integral. I looked in the table and found a similar formula 
. But I have a complex argument and I can’t immediately use the formula. However, if I manage to get it under the differential sign, then everything will be fine. If I write it down, then. But in the original integral there is no factor-three, therefore, in order for the integrand function not to change, I need to multiply it by ". In the course of approximately such mental reasoning, the following entry is born:
Now you can use the tabular formula 
:
Ready
The only difference is that we do not have the letter “X”, but a complex expression.
Let's check. Open the table of derivatives and differentiate the answer: 
The original integrand function has been obtained, which means that the integral has been found correctly.
Please note that during the verification we used the rule for differentiating a complex function 
. In essence, subsuming the function under the differential sign and 
- these are two mutually inverse rules.
Example 2
Let's analyze the integrand function. Here we have a fraction, and the denominator is a linear function (with “x” to the first power). We look at the table of integrals and find the most similar thing: 
.
We bring the function under the differential sign: 
Those who find it difficult to immediately figure out which fraction to multiply by can quickly reveal the differential in a draft: . Yeah, it turns out that this means that in order for nothing to change, I need to multiply the integral by .
Next we use the tabular formula 
:
Examination: 
The original integrand function has been obtained, which means that the integral has been found correctly.
Example 3
Find the indefinite integral. Perform check.
Example 4
Find the indefinite integral. Perform check.
This is an example for you to solve on your own. The answer is at the end of the lesson.
With some experience in solving integrals, such examples will seem easy and click like nuts:
At the end of this section, I would also like to dwell on the “free” case, when in a linear function a variable enters with a unit coefficient, for example:
Strictly speaking, the solution should look like this:
As you can see, subsuming the function under the differential sign was “painless”, without any multiplications. Therefore, in practice, such a long solution is often neglected and immediately written down that 
. But be prepared, if necessary, to explain to the teacher how you solved it! Because there is actually no integral in the table.
Variable change method in indefinite integral
Let's move on to consider the general case - the method of changing variables in the indefinite integral.
Example 5
Find the indefinite integral.
As an example, I took the integral that we looked at at the very beginning of the lesson. As we have already said, to solve the integral we liked the tabular formula 
, and I would like to reduce the whole matter to her.
The idea behind the replacement method is to replace a complex expression (or some function) with a single letter.
In this case it begs:
The second most popular replacement letter is the letter .
In principle, you can use other letters, but we will still adhere to traditions.
So: 
But when we replace it, we are left with ! Probably, many have guessed that if a transition is made to a new variable, then in the new integral everything should be expressed through the letter, and there is no place for a differential there at all.
The logical conclusion is that it is necessary turn into some expression that depends only on.
The action is as follows. After we have selected a replacement, in this example, we need to find the differential. With differentials, I think everyone has already established friendship.
Since then
After disassembling the differential, I recommend rewriting the final result as briefly as possible:
Now, according to the rules of proportion, we express what we need:
Eventually: 
Thus: 
And this is already the most tabular integral 
(the table of integrals, of course, is also valid for the variable).
Finally, all that remains is to carry out the reverse replacement. Let us remember that.
Ready.
The final design of the example considered should look something like this:
“
Let's replace: 
“
The icon does not have any mathematical meaning; it means that we have interrupted the solution for intermediate explanations.
When preparing an example in a notebook, it is better to mark the reverse substitution with a simple pencil.
Attention! In the following examples, finding the differential will not be described in detail.
And now it’s time to remember the first solution: 
What is the difference? There is no fundamental difference. It's actually the same thing. But from the point of view of designing the task, the method of subsuming a function under the differential sign is much shorter.
The question arises. If the first method is shorter, then why use the replacement method? The fact is that for a number of integrals it is not so easy to “fit” the function to the sign of the differential.
Example 6
Find the indefinite integral. 
Let's make a replacement: (it's hard to think of another replacement here) 
As you can see, as a result of the replacement, the original integral was significantly simplified - reduced to an ordinary power function. This is the purpose of the replacement - to simplify the integral.
Lazy advanced people can easily solve this integral by subsuming the function under the differential sign: 
Another thing is that such a solution is obviously not for all students. In addition, already in this example, the use of the method of subsuming a function under the differential sign significantly increases the risk of getting confused in a decision.
Example 7
Find the indefinite integral. Perform check.
Example 8
Find the indefinite integral. 
Replacement:
It remains to be seen what it will turn into 
Okay, we have expressed it, but what to do with the “X” remaining in the numerator?!
From time to time, when solving integrals, we encounter the following trick: we will express from the same replacement ! 
Example 9
Find the indefinite integral.
This is an example for you to solve on your own. The answer is at the end of the lesson.
Example 10
Find the indefinite integral.
Surely some people noticed that in my lookup table there is no variable replacement rule. This was done deliberately. The rule would create confusion in explanation and understanding, since it does not appear explicitly in the above examples.
Now it's time to talk about the basic premise of using the variable substitution method: the integrand must contain some function and its derivative:(functions may not be in the product)
In this regard, when finding integrals, you often have to look at the table of derivatives.
In the example under consideration, we notice that the degree of the numerator is one less than the degree of the denominator. In the table of derivatives we find the formula, which just reduces the degree by one. And that means that if you designate it as the denominator, then the chances are high that the numerator will turn into something good.
