Table of basic formulas for the definite integral. Antiderivative

On this page you will find:

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

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 it is necessary 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 looked at 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 “right away” using any standard rules. Moreover, the bad news is that, unlike the derivative, the antiderivative may not be considered at all. If we write absolutely random function and we 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 also 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 start, as always, with a repetition: let’s remember what an antiderivative is, why there are infinitely many of them and how to define them general form. 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

Here also we're talking about O trigonometric functions. 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 count the antiderivatives simple functions, you need to learn the 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 very important point: antiderivatives, like derivatives, can be considered a set in various ways. However, if all calculations and calculations are equal, then the answer will be the same. We have just seen this with 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 more powerful and useful tool, rather than simple “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 in this case it's quite simple to do this:

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 previous functions there is no variable $x$ above at all, i.e. We don’t understand what to add or subtract to get at least something similar to what is below. However, in fact, this expression is considered even simpler than any expression from the previous constructions, 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 on all kinds of tests and tests. 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 very important rule, which I initially did not plan to discuss at all in today’s video tutorial, 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 powers, we would get such a volume of calculations that simple task would borrow from us inadequately a large number 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.

The school algebra course includes integration and differentiation. To study this material you need tables of derivatives and integrals. In order to understand how to use them, you need to define the basic terms.

Derivative f(x) – characteristic of the intensity of change in the antiderivative function F(x) at any point on the graph. It expresses the limiting ratio of the increments of a function and its argument, which tends to zero. If a function has a finite derivative at any point, then it is differentiable. Calculating the derivative is differentiation.

Integral∫ is the inverse of the derivative, which expresses the size of the area of a certain part of the graph. The process of integration is finding the antiderivative function.

The same function can have several antiderivatives. For example, x^2. The main antiderivatives for it are x^3/3; x^3/3+1. The last digit is denoted by the letter C and the formula is as follows:

If C represents an arbitrary value, the integral is indefinite, if specific, it is definite.

Tables of derivative functions and integral tables will help you quickly and correctly cope with complex mathematical tasks. They include the most commonly used values, so students don't have to memorize a lot of formulas.

Table of derivative functions

To necessary materials were always at hand, you can download a table of derivative formulas . It contains formulas for calculating derivatives of basic elementary functions:

trigonometric;
logarithmic;
sedate;
exponential.

In addition, there is a special derivative table complex functions . It also contains formulas for the product of functions, their sum and quotient.

Table of indefinite and definite integrals

To quickly and correctly complete integration tasks, you can download tables of integrals, which contains all the most used formulas. They consist of two columns: the first contains mathematical formulas, the second is written explanations.

The tables include basic integrals the following functions:

rational;
exponential;
logarithmic;
irrational;
trigonometric;
hyperbolic.

In addition, you can download a table of indefinite integrals.

Cheat sheets with tables of integrals and derivatives

Many teachers require students to memorize complex formulas. The easiest way to memorize is constant practice, and to ensure that the necessary materials are at hand, you need to print them out.

Cheat sheet with derivative tables and integrals will help you quickly remember all the necessary formulas and successfully pass exams. To make it compact and easy to use, you need to choose the A5 format - half a regular sheet.

In an earlier material, the issue of finding the derivative was considered and its various applications: calculating the angular coefficient of a tangent to a graph, solving optimization problems, studying functions for monotonicity and extrema. $\newcommand(\tg)(\mathop(\mathrm(tg))\nolimits)$ $\newcommand(\ctg)(\mathop(\mathrm(tg))\nolimits)$ $\newcommand(\arctg)( \mathop(\mathrm(arctg))\nolimits)$ $\newcommand(\arcctg)(\mathop(\mathrm(arcctg))\nolimits)$

Picture 1.

The problem of finding the instantaneous velocity $v(t)$ using the derivative along a previously known path traveled, expressed by the function $s(t)$, was also considered.

Figure 2.

The inverse problem is also very common, when you need to find the path $s(t)$ traversed by a point in time $t$, knowing the speed of the point $v(t)$. If we recall, the instantaneous speed $v(t)$ is found as the derivative of the path function $s(t)$: $v(t)=s’(t)$. This means that in order to solve the inverse problem, that is, calculate the path, you need to find a function whose derivative will be equal to the speed function. But we know that the derivative of the path is the speed, that is: $s’(t) = v(t)$. Velocity is equal to acceleration times time: $v=at$. It is easy to determine that the desired path function will have the form: $s(t) = \frac(at^2)(2)$. But this is not quite a complete solution. Complete solution will have the form: $s(t)= \frac(at^2)(2)+C$, where $C$ is some constant. Why this is so will be discussed further. For now, let's check the correctness of the solution found: $s"(t)=\left(\frac(at^2)(2)+C\right)"=2\frac(at)(2)+0=at=v( t)$.

It is worth noting that finding a path by speed is physical meaning antiderivative.

The resulting function $s(t)$ is called the antiderivative of the function $v(t)$. Quite interesting and unusual name, is not it. It contains a lot of meaning that explains the essence this concept and leads to its understanding. You will notice that it contains two words “first” and “image”. They speak for themselves. That is, this is the function that is the initial one for the derivative we have. And using this derivative we are looking for the function that was at the beginning, was “first”, “first image”, that is, antiderivative. It is sometimes also called a primitive function or antiderivative.

As we already know, the process of finding the derivative is called differentiation. And the process of finding the antiderivative is called integration. The operation of integration is the inverse of the operation of differentiation. The converse is also true.

Definition. An antiderivative for a function $f(x)$ on a certain interval is a function $F(x)$ whose derivative is equal to this function $f(x)$ for all $x$ from the specified interval: $F'(x)=f (x)$.

Someone may have a question: where did $F(x)$ and $f(x)$ come from in the definition, if initially we were talking about $s(t)$ and $v(t)$. The fact is that $s(t)$ and $v(t)$ are special cases of function designation that have a specific meaning in this case, that is, they are a function of time and a function of speed, respectively. It's the same with the variable $t$ - it denotes time. And $f$ and $x$ are the traditional variant of the general designation of a function and a variable, respectively. Worth paying Special attention to the designation of the antiderivative $F(x)$. First of all, $F$ is capital. Antiderivatives are designated in capital letters. Secondly, the letters are the same: $F$ and $f$. That is, for the function $g(x)$ the antiderivative will be denoted by $G(x)$, for $z(x)$ – by $Z(x)$. Regardless of the notation, the rules for finding an antiderivative function are always the same.

Let's look at a few examples.

Example 1. Prove that the function $F(x)=\frac(1)(5)\sin5x$ is an antiderivative of the function $f(x)=\cos5x$.

To prove this, we will use the definition, or rather the fact that $F'(x)=f(x)$, and find the derivative of the function $F(x)$: $F'(x)=(\frac(1)(5 ) \sin5x)'=\frac(1)(5)\cdot 5\cos5x= \cos5x$. This means $F(x)=\frac(1)(5) \sin5x$ is the antiderivative of $f(x)=\cos5x$. Q.E.D.

Example 2. Find which functions correspond to the following antiderivatives: a) $F(z)=\tg z$; b) $G(l) = \sin l$.

To find the required functions, let's calculate their derivatives:
a) $F’(z)=(\tg z)’=\frac(1)(\cos^2 z)$;
b) $G(l) = (\sin l)’ = \cos l$.

Example 3. What will be the antiderivative for $f(x)=0$?
Let's use the definition. Let's think about which function can have a derivative equal to $0$. Recalling the table of derivatives, we find that any constant will have such a derivative. We find that the antiderivative we are looking for is: $F(x)= C$.

The resulting solution can be explained geometrically and physically. Geometrically, it means that the tangent to the graph $y=F(x)$ is horizontal at each point of this graph and, therefore, coincides with the $Ox$ axis. Physically it is explained by the fact that a point having a speed equal to zero, remains in place, that is, the path it has traveled remains unchanged. Based on this, we can formulate the following theorem.

Theorem. (Sign of constancy of functions). If on some interval $F’(x) = 0$, then the function $F(x)$ on this interval is constant.

Example 4. Determine which functions are antiderivatives of a) $F_1 = \frac(x^7)(7)$; b) $F_2 = \frac(x^7)(7) – 3$; c) $F_3 = \frac(x^7)(7) + 9$; d) $F_4 = \frac(x^7)(7) + a$, where $a$ is some number.
Using the definition of an antiderivative, we conclude that to solve this problem we need to calculate the derivatives of the data we have in the original different functions. When calculating, remember that the derivative of a constant, that is, of any number, is equal to zero.
a) $F_1 =(\frac(x^7)(7))"= 7 \cdot \frac(x^6)(7) = x^6$;
b) $F_2 =\left(\frac(x^7)(7) – 3\right)"=7 \cdot \frac(x^6)(7)= x^6$;
c) $F_3 =(\frac(x^7)(7) + 9)’= x^6$;
d) $F_4 =(\frac(x^7)(7) + a)’ = x^6$.

What do we see? Several different functions are primitives of the same function. This suggests that any function has infinitely many antiderivatives, and they have the form $F(x) + C$, where $C$ is an arbitrary constant. That is, the operation of integration is multivalued, unlike the operation of differentiation. Based on this, let us formulate a theorem that describes the main property of antiderivatives.

Theorem. (The main property of antiderivatives). Let the functions $F_1$ and $F_2$ be antiderivatives of the function $f(x)$ on some interval. Then for all values from this interval the following equality is true: $F_2=F_1+C$, where $C$ is some constant.

The fact of the presence of an infinite number of antiderivatives can be interpreted geometrically. Using parallel translation along the $Oy$ axis, one can obtain from each other the graphs of any two antiderivatives for $f(x)$. This is the geometric meaning of the antiderivative.

It is very important to pay attention to the fact that by choosing the constant $C$ you can ensure that the graph of the antiderivative passes through a certain point.

Figure 3.

Example 5. Find an antiderivative for the function $f(x)=\frac(x^2)(3)+1$, the graph of which passes through the point $(3; 1)$.
Let's first find all antiderivatives for $f(x)$: $F(x)=\frac(x^3)(9)+x + C$.
Next, we will find a number C for which the graph $y=\frac(x^3)(9)+x + C$ will pass through the point $(3; 1)$. To do this, let’s substitute the coordinates of the point into the graph equation and solve it for $C$:
$1= \frac(3^3)(9)+3 + C$, $C=-5$.
We obtained a graph $y=\frac(x^3)(9)+x-5$, which corresponds to the antiderivative $F(x)=\frac(x^3)(9)+x-5$.

Table of antiderivatives

A table of formulas for finding antiderivatives can be compiled using formulas for finding derivatives.

Table of antiderivatives

Functions	Antiderivatives
$0$	$C$
$1$	$x+C$
$a\in R$	$ax+C$
$x^n, n\ne1$	$\displaystyle \frac(x^(n+1))(n+1)+C$
$\displaystyle \frac(1)(x)$	$\ln\|x\|+C$
$\sin x$	$-\cos x+C$
$\cos x$	$\sin x+C$
$\displaystyle \frac(1)(\sin^2 x)$	$-\ctg x+C$
$\displaystyle \frac(1)(\cos^2 x)$	$\tg x+C$
$e^x$	$e^x+C$
$a^x, a>0, a\ne1$	$\displaystyle \frac(a^x)(\ln a) +C$
$\displaystyle \frac(1)(\sqrt(1-x^2))$	$\arcsin x+C$
$\displaystyle -\frac(1)(\sqrt(1-x^2))$	$\arccos x+C$
$\displaystyle \frac(1)(1+x^2)$	$\arctg x+C$
$\displaystyle -\frac(1)(1+x^2)$	$\arcctg x+C$

You can check the correctness of the table in the following way: for each set of antiderivatives located in the right column, find the derivative, which will result in the corresponding functions in the left column.

Some rules for finding antiderivatives

As you know, many functions have more complex look, rather than those indicated in the table of antiderivatives, and can represent any arbitrary combination of sums and products of functions from this table. And here the question arises: how to calculate antiderivatives of such functions. For example, from the table we know how to calculate the antiderivatives $x^3$, $\sin x$ and $10$. How, for example, can one calculate the antiderivative $x^3-10\sin x$? Looking ahead, it is worth noting that it will be equal to $\frac(x^4)(4)+10\cos x$.
1. If $F(x)$ is antiderivative for $f(x)$, $G(x)$ for $g(x)$, then for $f(x)+g(x)$ the antiderivative will be equal to $ F(x)+G(x)$.
2. If $F(x)$ is an antiderivative for $f(x)$ and $a$ is a constant, then for $af(x)$ the antiderivative is $aF(x)$.
3. If for $f(x)$ the antiderivative is $F(x)$, $a$ and $b$ are constants, then $\frac(1)(a) F(ax+b)$ is the antiderivative for $f (ax+b)$.
Using the obtained rules we can expand the table of antiderivatives.

Functions	Antiderivatives
$(ax+b)^n, n\ne1, a\ne0$	$\displaystyle \frac((ax+b)^n)(a(n+1)) +C$
$\displaystyle \frac(1)(ax+b), a\ne0$	$\displaystyle \frac(1)(a)\ln\|ax+b\|+C$
$e^(ax+b), a\ne0$	$\displaystyle \frac(1)(a) e^(ax+b)+C$
$\sin(ax+b), a\ne0$	$\displaystyle -\frac(1)(a)\cos(ax+b)+C$
$\cos(ax+b), a\ne0$	$\displaystyle \frac(1)(a)\sin(ax+b)+C$