Integral of a complex power function. Calculation of the simplest indefinite integrals

Complex integrals

This article concludes the topic of indefinite integrals, and includes integrals that I find quite complex. The lesson was created at the repeated requests of visitors who expressed their wish that more difficult examples be analyzed on the site.

It is assumed that the reader of this text is well prepared and knows how to apply basic integration techniques. Dummies and people who are not very confident in integrals should refer to the very first lesson - Indefinite integral. Examples of solutions, where you can master the topic almost from scratch. More experienced students can become familiar with techniques and methods of integration that have not yet been encountered in my articles.

What integrals will be considered?

First we will consider integrals with roots, for the solution of which we successively use variable replacement And integration by parts. That is, in one example two techniques are combined at once. And even more.

Then we will get acquainted with interesting and original method of reducing the integral to itself. Quite a few integrals are solved this way.

The third issue of the program will be integrals of complex fractions, which flew past the cash desk in previous articles.

Fourthly, additional integrals from trigonometric functions will be analyzed. In particular, there are methods that avoid time-consuming universal trigonometric substitution.

(2) In the integrand function, we divide the numerator by the denominator term by term.

(3) We use the linearity property of the indefinite integral. In the last integral immediately put the function under the differential sign.

(4) We take the remaining integrals. Note that in a logarithm you can use parentheses rather than a modulus, since .

(5) We carry out a reverse replacement, expressing “te” from the direct replacement:

Masochistic students can differentiate the answer and get the original integrand, as I just did. No, no, I did the check in the right sense =)

As you can see, during the solution we had to use even more than two solution methods, so to deal with such integrals you need confident integration skills and quite a bit of experience.

In practice, of course, the square root is more common; here are three examples for solving it yourself:

Example 2

Find the indefinite integral

Example 3

Find the indefinite integral

Example 4

Find the indefinite integral

These examples are of the same type, so the complete solution at the end of the article will only be for Example 2; Examples 3-4 have the same answers. Which replacement to use at the beginning of decisions, I think, is obvious. Why did I choose examples of the same type? Often found in their role. More often, perhaps, just something like .

But not always, when under the arctangent, sine, cosine, exponential and other functions there is a root of a linear function, you have to use several methods at once. In a number of cases, it is possible to “get off easy,” that is, immediately after the replacement, a simple integral is obtained, which can be easily taken. The easiest of the tasks proposed above is Example 4, in which, after replacement, a relatively simple integral is obtained.

By reducing the integral to itself

A witty and beautiful method. Let's take a look at the classics of the genre:

Example 5

Find the indefinite integral

Under the root is a quadratic binomial, and trying to integrate this example can give the teapot a headache for hours. Such an integral is taken in parts and reduced to itself. In principle, it’s not difficult. If you know how.

Let us denote the integral under consideration by a Latin letter and begin the solution:

Let's integrate by parts:

(1) Prepare the integrand function for term-by-term division.

(2) We divide the integrand function term by term. It may not be clear to everyone, but I’ll describe it in more detail:

(3) We use the linearity property of the indefinite integral.

(4) Take the last integral (“long” logarithm).

Now let's look at the very beginning of the solution:

And to the end:

What happened? As a result of our manipulations, the integral was reduced to itself!

Let's equate the beginning and the end:

Move to the left side with a change of sign:

And we move the two to the right side. As a result:

The constant, strictly speaking, should have been added earlier, but I added it at the end. I strongly recommend reading what the rigor is here:

Note: More strictly, the final stage of the solution looks like this:

Thus:

The constant can be redesignated by . Why can it be redesignated? Because he still accepts it any values, and in this sense there is no difference between constants and.
As a result:

A similar trick with constant renotation is widely used in differential equations. And there I will be strict. And here I allow such freedom only in order not to confuse you with unnecessary things and to focus attention precisely on the integration method itself.

Example 6

Find the indefinite integral

Another typical integral for independent solution. Full solution and answer at the end of the lesson. There will be a difference with the answer in the previous example!

If under the square root there is a square trinomial, then the solution in any case comes down to two analyzed examples.

For example, consider the integral . All you need to do is first select a complete square:
.
Next, a linear replacement is carried out, which does “without any consequences”:
, resulting in the integral . Something familiar, right?

Or this example, with a quadratic binomial:
Select a complete square:
And, after linear replacement, we obtain the integral, which is also solved using the algorithm already discussed.

Let's look at two more typical examples of how to reduce an integral to itself:
– integral of the exponential multiplied by sine;
– integral of the exponential multiplied by the cosine.

In the listed integrals by parts you will have to integrate twice:

Example 7

Find the indefinite integral

The integrand is the exponential multiplied by the sine.

We integrate by parts twice and reduce the integral to itself:

As a result of double integration by parts, the integral was reduced to itself. We equate the beginning and end of the solution:

We move it to the left side with a change of sign and express our integral:

Ready. At the same time, it is advisable to comb the right side, i.e. take the exponent out of brackets, and place the sine and cosine in brackets in a “beautiful” order.

Now let's go back to the beginning of the example, or more precisely, to integration by parts:

We designated the exponent as. The question arises: is it the exponent that should always be denoted by ? Not necessary. In fact, in the considered integral fundamentally doesn't matter, what do we mean by , we could have gone the other way:

Why is this possible? Because the exponential turns into itself (both during differentiation and integration), sine and cosine mutually turn into each other (again, both during differentiation and integration).

That is, we can also denote a trigonometric function. But, in the example considered, this is less rational, since fractions will appear. If you wish, you can try to solve this example using the second method; the answers must match.

Example 8

Find the indefinite integral

This is an example for you to solve on your own. Before you decide, think about what is more advantageous in this case to designate as , an exponential or a trigonometric function? Full solution and answer at the end of the lesson.

And, of course, do not forget that most of the answers in this lesson are quite easy to check by differentiation!

The examples considered were not the most complex. In practice, integrals are more common where the constant is both in the exponent and in the argument of the trigonometric function, for example: . Many people will get confused in such an integral, and I often get confused myself. The fact is that there is a high probability of fractions appearing in the solution, and it is very easy to lose something through carelessness. In addition, there is a high probability of an error in the signs; note that the exponent has a minus sign, and this introduces additional difficulty.

At the final stage, the result is often something like this:

Even at the end of the solution, you should be extremely careful and correctly understand the fractions:

Integrating Complex Fractions

We are slowly approaching the equator of the lesson and begin to consider integrals of fractions. Again, not all of them are super complex, it’s just that for one reason or another the examples were a little “off topic” in other articles.

Continuing the theme of roots

Example 9

Find the indefinite integral

In the denominator under the root there is a quadratic trinomial plus an “appendage” in the form of an “X” outside the root. An integral of this type can be solved using a standard substitution.

We decide:

The replacement here is simple:

Let's look at life after replacement:

(1) After substitution, we reduce the terms under the root to a common denominator.
(2) We take it out from under the root.
(3) The numerator and denominator are reduced by . At the same time, under the root, I rearranged the terms in a convenient order. With some experience, steps (1), (2) can be skipped by performing the commented actions orally.
(4) The resulting integral, as you remember from the lesson Integrating Some Fractions, is being decided complete square extraction method. Select a complete square.
(5) By integration we obtain an ordinary “long” logarithm.
(6) We carry out the reverse replacement. If initially , then back: .
(7) The final action is aimed at straightening the result: under the root we again bring the terms to a common denominator and take them out from under the root.

Example 10

Find the indefinite integral

This is an example for you to solve on your own. Here a constant is added to the lone “X”, and the replacement is almost the same:

The only thing you need to do additionally is to express the “x” from the replacement being carried out:

Full solution and answer at the end of the lesson.

Sometimes in such an integral there may be a quadratic binomial under the root, this does not change the method of solution, it will be even simpler. Feel the difference:

Example 11

Find the indefinite integral

Example 12

Find the indefinite integral

Brief solutions and answers at the end of the lesson. It should be noted that Example 11 is exactly binomial integral, the solution method of which was discussed in class Integrals of irrational functions.

Integral of an indecomposable polynomial of the 2nd degree to the power

(polynomial in denominator)

A more rare type of integral, but nevertheless encountered in practical examples.

Example 13

Find the indefinite integral

But let’s return to the example with lucky number 13 (honestly, I didn’t guess correctly). This integral is also one of those that can be quite frustrating if you don’t know how to solve.

The solution starts with an artificial transformation:

I think everyone already understands how to divide the numerator by the denominator term by term.

The resulting integral is taken in parts:

For an integral of the form ( – natural number) we derive recurrent reduction formula:
, Where – integral of a degree lower.

Let us verify the validity of this formula for the solved integral.
In this case: , , we use the formula:

As you can see, the answers are the same.

Example 14

Find the indefinite integral

This is an example for you to solve on your own. The sample solution uses the above formula twice in succession.

If under the degree is indivisible square trinomial, then the solution is reduced to a binomial by isolating the perfect square, for example:

What if there is an additional polynomial in the numerator? In this case, the method of indefinite coefficients is used, and the integrand function is expanded into a sum of fractions. But in my practice there is such an example never met, so I missed this case in the article Integrals of fractional-rational functions, I'll skip it now. If you still encounter such an integral, look at the textbook - everything is simple there. I don’t think it’s advisable to include material (even simple ones), the probability of encountering which tends to zero.

Integrating complex trigonometric functions

The adjective “complex” for most examples is again largely conditional. Let's start with tangents and cotangents in high powers. From the point of view of the solving methods used, tangent and cotangent are almost the same thing, so I will talk more about tangent, implying that the demonstrated method for solving the integral is valid for cotangent too.

In the above lesson we looked at universal trigonometric substitution for solving a certain type of integrals of trigonometric functions. The disadvantage of universal trigonometric substitution is that its use often results in cumbersome integrals with difficult calculations. And in some cases, universal trigonometric substitution can be avoided!

Let's consider another canonical example, the integral of one divided by sine:

Example 17

Find the indefinite integral

Here you can use universal trigonometric substitution and get the answer, but there is a more rational way. I will provide the complete solution with comments for each step:

(1) We use the trigonometric formula for the sine of a double angle.
(2) We carry out an artificial transformation: Divide in the denominator and multiply by .
(3) Using the well-known formula in the denominator, we transform the fraction into a tangent.
(4) We bring the function under the differential sign.
(5) Take the integral.

A couple of simple examples for you to solve on your own:

Example 18

Find the indefinite integral

Note: The very first step should be to use the reduction formula and carefully carry out actions similar to the previous example.

Example 19

Find the indefinite integral

Well, this is a very simple example.

Complete solutions and answers at the end of the lesson.

I think now no one will have problems with integrals:
and so on.

What is the idea of the method? The idea is to use transformations and trigonometric formulas to organize only tangents and the tangent derivative into the integrand. That is, we are talking about replacing: . In Examples 17-19 we actually used this replacement, but the integrals were so simple that we got by with an equivalent action - subsuming the function under the differential sign.

Similar reasoning, as I already mentioned, can be carried out for the cotangent.

There is also a formal prerequisite for applying the above replacement:

The sum of the powers of cosine and sine is a negative integer EVEN number, For example:

for the integral – a negative integer EVEN number.

! Note : if the integrand contains ONLY a sine or ONLY a cosine, then the integral is also taken for a negative odd degree (the simplest cases are in Examples No. 17, 18).

Let's look at a couple of more meaningful tasks based on this rule:

Example 20

Find the indefinite integral

The sum of the powers of sine and cosine: 2 – 6 = –4 is a negative integer EVEN number, which means that the integral can be reduced to tangents and its derivative:

(1) Let's transform the denominator.
(2) Using the well-known formula, we obtain .
(3) Let's transform the denominator.
(4) We use the formula .
(5) We bring the function under the differential sign.
(6) We carry out replacement. More experienced students may not carry out the replacement, but it is still better to replace the tangent with one letter - there is less risk of getting confused.

Example 21

Find the indefinite integral

This is an example for you to solve on your own.

Hang in there, the championship rounds are about to begin =)

Often the integrand contains a “hodgepodge”:

Example 22

Find the indefinite integral

This integral initially contains a tangent, which immediately leads to an already familiar thought:

I will leave the artificial transformation at the very beginning and the remaining steps without comment, since everything has already been discussed above.

A couple of creative examples for your own solution:

Example 23

Find the indefinite integral

Example 24

Find the indefinite integral

Yes, in them, of course, you can lower the powers of sine and cosine, and use a universal trigonometric substitution, but the solution will be much more efficient and shorter if it is carried out through tangents. Full solution and answers at the end of the lesson

It is shown that the integral of the product of power functions of sin x and cos x can be reduced to the integral of a differential binomial. For integer values of exponents, such integrals are easily calculated by parts or using reduction formulas. The derivation of the reduction formulas is given. An example of calculating such an integral is given.

Content

Reduction to the integral of a differential binomial

Let's consider integrals of the form:

Such integrals are reduced to the integral of the differential binomial of one of the substitutions t = sin x or t = cos x.

Let's demonstrate this by performing the substitution
t = sin x.
Then
dt = (sin x)′ dx = cos x dx;
cos 2 x = 1 - sin 2 x = 1 - t 2;

If m and n are rational numbers, then differential binomial integration methods should be used.

Integration with integers m and n

Next, consider the case when m and n are integers (not necessarily positive). In this case, the integrand is a rational function of sin x And cos x. Therefore, you can apply the rules presented in the section "Integrating trigonometric rational functions".

However, taking into account the specific features, it is easier to use reduction formulas, which are easily obtained by integration by parts.

Reduction formulas

Reduction formulas for the integral

have the form:

;
;
;
.

There is no need to memorize them, since they are easily obtained by integrating by parts.

Proof of reduction formulas

Let's integrate by parts.

Multiplying by m + n, we get the first formula:

We similarly obtain the second formula.

Let's integrate by parts.

Multiplying by m + n, we get the second formula:

Third formula.

Let's integrate by parts.

Multiplying by n + 1 , we get the third formula:

Similarly, for the fourth formula.

Let's integrate by parts.

Multiplying by m + 1 , we get the fourth formula:

Example

Let's calculate the integral:

Let's transform:

Here m = 10, n = - 4.

We apply the reduction formula:

When m = 10, n = - 4:

When m = 8, n = - 2:

We apply the reduction formula:

When m = 6, n = - 0:

When m = 4, n = - 0:

When m = 2, n = - 0:

We calculate the remaining integral:

We collect intermediate results into one formula.

References:
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003.

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:

Problem 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.

Using a "secret" technique

In conclusion, I would like to look at another interesting technique, which, on the one hand, goes beyond 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 do the ninth degree at all 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.

Integral of a complex power function. Calculation of the simplest indefinite integrals

Complex integrals

By reducing the integral to itself

Integrating Complex Fractions

Integral of an indecomposable polynomial of the 2nd degree to the power

Integrating complex trigonometric functions

Reduction to the integral of a differential binomial

Integration with integers m and n

Reduction formulas

Proof of reduction formulas

Example

Solving easy examples

Example #1

Example No. 2

Solving problems containing an exponential function

Example #1

Example No. 2

Rules for working with the table of antiderivatives

Problem solving: finding the antiderivative of a function

Example #1

Example No. 2

Nuances of the solution

We train in integration in practice

Task No. 1

Problem No. 2

Problem No. 3

More complex problems for solving antiderivative exponential functions

Task No. 1

Problem No. 2

Using a "secret" technique

Task No. 1

Problem No. 2

Nuances of the solution