Principal integrals that every student should know
The listed integrals are the basis, the basis of the fundamentals. These formulas should definitely be remembered. When calculating more complex integrals, you will have to use them constantly.
Pay special attention to formulas (5), (7), (9), (12), (13), (17) and (19). Don't forget to add an arbitrary constant C to your answer when integrating!
Integral of a constant
∫ A d x = A x + C (1)Integrating a Power Function
In fact, it was possible to limit ourselves to only formulas (5) and (7), but the rest of the integrals from this group occur so often that it is worth paying a little attention to them.
∫ x d x = x 2 2 + C (2)
∫ x 2 d x = x 3 3 + C (3)
∫ 1 x d x = 2 x + C (4)
∫ 1 x d x = ln | x | +C (5)
∫ 1 x 2 d x = − 1 x + C (6)
∫ x n d x = x n + 1 n + 1 + C (n ≠ − 1) (7)
Integrals of exponential functions and hyperbolic functions
Of course, formula (8) (perhaps the most convenient for memorization) can be considered as a special case of formula (9). Formulas (10) and (11) for the integrals of the hyperbolic sine and hyperbolic cosine are easily derived from formula (8), but it is better to simply remember these relations.
∫ e x d x = e x + C (8)
∫ a x d x = a x ln a + C (a > 0, a ≠ 1) (9)
∫ s h x d x = c h x + C (10)
∫ c h x d x = s h x + C (11)
Basic integrals of trigonometric functions
A mistake that students often make is that they confuse the signs in formulas (12) and (13). Remembering that the derivative of the sine is equal to the cosine, for some reason many people believe that the integral of the function sinx is equal to cosx. This is not true! The integral of sine is equal to “minus cosine”, but the integral of cosx is equal to “just sine”:
∫ sin x d x = − cos x + C (12)
∫ cos x d x = sin x + C (13)
∫ 1 cos 2 x d x = t g x + C (14)
∫ 1 sin 2 x d x = − c t g x + C (15)
Integrals that reduce to inverse trigonometric functions
Formula (16), leading to the arctangent, is naturally a special case of formula (17) for a=1. Similarly, (18) is a special case of (19).
∫ 1 1 + x 2 d x = a r c t g x + C = − a r c c t g x + C (16)
∫ 1 x 2 + a 2 = 1 a a r c t g x a + C (a ≠ 0) (17)
∫ 1 1 − x 2 d x = arcsin x + C = − arccos x + C (18)
∫ 1 a 2 − x 2 d x = arcsin x a + C = − arccos x a + C (a > 0) (19)
More complex integrals
It is also advisable to remember these formulas. They are also used quite often, and their output is quite tedious.
∫ 1 x 2 + a 2 d x = ln | x + x 2 + a 2 | +C (20)
∫ 1 x 2 − a 2 d x = ln | x + x 2 − a 2 | +C (21)
∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 arcsin x a + C (a > 0) (22)
∫ x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 ln | x + x 2 + a 2 | + C (a > 0) (23)
∫ x 2 − a 2 d x = x 2 x 2 − a 2 − a 2 2 ln | x + x 2 − a 2 | + C (a > 0) (24)
General rules of integration
1) The integral of the sum of two functions is equal to the sum of the corresponding integrals: ∫ (f (x) + g (x)) d x = ∫ f (x) d x + ∫ g (x) d x (25)
2) The integral of the difference of two functions is equal to the difference of the corresponding integrals: ∫ (f (x) − g (x)) d x = ∫ f (x) d x − ∫ g (x) d x (26)
3) The constant can be taken out of the integral sign: ∫ C f (x) d x = C ∫ f (x) d x (27)
It is easy to see that property (26) is simply a combination of properties (25) and (27).
4) Integral of a complex function if the internal function is linear: ∫ f (A x + B) d x = 1 A F (A x + B) + C (A ≠ 0) (28)
Here F(x) is an antiderivative for the function f(x). Please note: this formula only works when the inner function is Ax + B.
Important: there is no universal formula for the integral of the product of two functions, as well as for the integral of a fraction:
∫ f (x) g (x) d x = ? ∫ f (x) g (x) d x = ? (thirty)
This does not mean, of course, that a fraction or product cannot be integrated. It’s just that every time you see an integral like (30), you will have to invent a way to “fight” it. In some cases, integration by parts will help you, in others you will have to make a change of variable, and sometimes even “school” algebra or trigonometry formulas can help.
A simple example of calculating the indefinite integral
Example 1. Find the integral: ∫ (3 x 2 + 2 sin x − 7 e x + 12) d xLet us use formulas (25) and (26) (the integral of the sum or difference of functions is equal to the sum or difference of the corresponding integrals. We obtain: ∫ 3 x 2 d x + ∫ 2 sin x d x − ∫ 7 e x d x + ∫ 12 d x
Let us remember that the constant can be taken out of the integral sign (formula (27)). The expression is converted to the form
3 ∫ x 2 d x + 2 ∫ sin x d x − 7 ∫ e x d x + 12 ∫ 1 d x
Now let's just use the table of basic integrals. We will need to apply formulas (3), (12), (8) and (1). Let's integrate the power function, sine, exponential and constant 1. Don't forget to add an arbitrary constant C at the end:
3 x 3 3 − 2 cos x − 7 e x + 12 x + C
After elementary transformations we get the final answer:
X 3 − 2 cos x − 7 e x + 12 x + C
Test yourself by differentiation: take the derivative of the resulting function and make sure that it is equal to the original integrand.
Summary table of integrals
| ∫ A d x = A x + C |
| ∫ x d x = x 2 2 + C |
| ∫ x 2 d x = x 3 3 + C |
| ∫ 1 x d x = 2 x + C |
| ∫ 1 x d x = ln | x | +C |
| ∫ 1 x 2 d x = − 1 x + C |
| ∫ x n d x = x n + 1 n + 1 + C (n ≠ − 1) |
| ∫ e x d x = e x + C |
| ∫ a x d x = a x ln a + C (a > 0, a ≠ 1) |
| ∫ s h x d x = c h x + C |
| ∫ c h x d x = s h x + C |
| ∫ sin x d x = − cos x + C |
| ∫ cos x d x = sin x + C |
| ∫ 1 cos 2 x d x = t g x + C |
| ∫ 1 sin 2 x d x = − c t g x + C |
| ∫ 1 1 + x 2 d x = a r c t g x + C = − a r c c t g x + C |
| ∫ 1 x 2 + a 2 = 1 a a r c t g x a + C (a ≠ 0) |
| ∫ 1 1 − x 2 d x = arcsin x + C = − arccos x + C |
| ∫ 1 a 2 − x 2 d x = arcsin x a + C = − arccos x a + C (a > 0) |
| ∫ 1 x 2 + a 2 d x = ln | x + x 2 + a 2 | +C |
| ∫ 1 x 2 − a 2 d x = ln | x + x 2 − a 2 | +C |
| ∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 arcsin x a + C (a > 0) |
| ∫ x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 ln | x + x 2 + a 2 | + C (a > 0) |
| ∫ x 2 − a 2 d x = x 2 x 2 − a 2 − a 2 2 ln | x + x 2 − a 2 | + C (a > 0) |
Download the table of integrals (part II) from this link
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Let us list the integrals of elementary functions, which are sometimes called tabular:
Any of the above formulas can be proven by taking the derivative of the right-hand side (the result will be the integrand).
Integration methods
Let's look at some basic integration methods. These include:
1. Decomposition method(direct integration).
This method is based on the direct use of tabular integrals, as well as on the use of properties 4 and 5 of the indefinite integral (i.e., taking the constant factor out of brackets and/or representing the integrand as a sum of functions - decomposition of the integrand into terms).
Example 1. For example, to find(dx/x 4) you can directly use the table integral forx n dx. In fact,(dx/x 4) =x -4 dx=x -3 /(-3) +C= -1/3x 3 +C.
Let's look at a few more examples.
Example 2. To find it, we use the same integral:
Example 3. To find it you need to take
Example 4. To find, we represent the integrand function in the form 
and use the table integral for the exponential function:
Let's consider the use of bracketing a constant factor.
Example 5.
Let's find, for example 
. Considering that, we get
Example 6. We'll find it. Because the 
, let's use the table integral 
We get
In the following two examples, you can also use bracketing and table integrals:
Example 7.
(we use and 
);
Example 8.
(we use 
And 
).
Let's look at more complex examples that use the sum integral.
Example 9. For example, let's find 
. To apply the expansion method in the numerator, we use the sum cube formula , and then divide the resulting polynomial by the denominator, term by term.
=((8x 3/2 + 12x+ 6x 1/2 + 1)/(x 3/2))dx=(8 + 12x -1/2 + 6/x+x -3/2)dx= 8 dx+ 12x -1/2 dx+ + 6dx/x+x -3/2 dx= 
It should be noted that at the end of the solution one common constant C is written (and not separate ones when integrating each term). In the future, it is also proposed to omit the constants from the integration of individual terms in the solution process as long as the expression contains at least one indefinite integral (we will write one constant at the end of the solution).
Example 10. We'll find 
. To solve this problem, let's factorize the numerator (after this we can reduce the denominator).
Example 11. We'll find it. Trigonometric identities can be used here.
Sometimes, in order to decompose an expression into terms, you have to use more complex techniques.
Example 12. We'll find 
. In the integrand we select the whole part of the fraction 
. Then
Example 13. We'll find
2. Variable replacement method (substitution method)
The method is based on the following formula: f(x)dx=f((t))`(t)dt, where x =(t) is a function differentiable on the interval under consideration.
Proof. Let's find the derivatives with respect to the variable t from the left and right sides of the formula.
Note that on the left side there is a complex function whose intermediate argument is x = (t). Therefore, to differentiate it with respect to t, we first differentiate the integral with respect to x, and then take the derivative of the intermediate argument with respect to t.
( f(x)dx)` t = ( f(x)dx)` x *x` t = f(x) `(t)
Derivative from the right side:
(f((t))`(t)dt)` t =f((t))`(t) =f(x)`(t)
Since these derivatives are equal, by corollary to Lagrange’s theorem, the left and right sides of the formula being proved differ by a certain constant. Since the indefinite integrals themselves are defined up to an indefinite constant term, this constant can be omitted from the final notation. Proven.
A successful change of variable allows you to simplify the original integral, and in the simplest cases, reduce it to a tabular one. In the application of this method, a distinction is made between linear and nonlinear substitution methods.
a) Linear substitution method Let's look at an example.
Example 1.
. Let t= 1 – 2x, then
dx=d(½ - ½t) = - ½dt
It should be noted that the new variable does not need to be written out explicitly. In such cases, they talk about transforming a function under the differential sign or about introducing constants and variables under the differential sign, i.e. O implicit variable replacement.
Example 2. For example, let's findcos(3x + 2)dx. By the properties of the differential dx = (1/3)d(3x) = (1/3)d(3x + 2), thencos(3x + 2)dx =(1/3)cos(3x + 2)d (3x + + 2) = (1/3)cos(3x + 2)d(3x + 2) = (1/3)sin(3x + 2) +C.
In both examples considered, linear substitution t=kx+b(k0) was used to find the integrals.
In the general case, the following theorem is valid.
Linear substitution theorem. Let F(x) be some antiderivative of the function f(x). Thenf(kx+b)dx= (1/k)F(kx+b) +C, where k and b are some constants,k0.
Proof.
By definition of the integral f(kx+b)d(kx+b) =F(kx+b) +C. Hod(kx+b)= (kx+b)`dx=kdx. Let's take the constant factor k out of the integral sign: kf(kx+b)dx=F(kx+b) +C. Now we can divide the left and right sides of the equality into two and obtain the statement to be proved up to the designation of the constant term.
This theorem states that if in the definition of the integral f(x)dx= F(x) + C instead of the argument x we substitute the expression (kx+b), this will lead to the appearance of an additional factor 1/k in front of the antiderivative.
Using the proven theorem, we solve the following examples.
Example 3.
We'll find 
. Here kx+b= 3 –x, i.e. k= -1,b= 3. Then
Example 4.
We'll find it. Herekx+b= 4x+ 3, i.e. k= 4,b= 3. Then
Example 5.
We'll find 
. Here kx+b= -2x+ 7, i.e. k= -2,b= 7. Then
.
Example 6. We'll find 
. Here kx+b= 2x+ 0, i.e. k= 2,b= 0.
.
Let us compare the result obtained with example 8, which was solved by the decomposition method. Solving the same problem using a different method, we got the answer 
. Let's compare the results: Thus, these expressions differ from each other by a constant term 
, i.e. The answers received do not contradict each other.
Example 7. We'll find 
. Let's select a perfect square in the denominator.
In some cases, changing a variable does not reduce the integral directly to a tabular one, but can simplify the solution, making it possible to use the expansion method at a subsequent step.
Example 8. For example, let's find 
. Replace t=x+ 2, then dt=d(x+ 2) =dx. Then
,
where C = C 1 – 6 (when substituting the expression (x+ 2) instead of the first two terms we get ½x 2 -2x– 6).
Example 9. We'll find 
. Let t= 2x+ 1, then dt= 2dx;dx= ½dt;x= (t– 1)/2.
Let's substitute the expression (2x+ 1) for t, open the brackets and give similar ones.
Note that in the process of transformations we moved to another constant term, because the group of constant terms could be omitted during the transformation process.
b) Nonlinear substitution method Let's look at an example.
Example 1.
. Lett= -x 2. Next, one could express x in terms of t, then find an expression for dx and implement a change of variable in the desired integral. But in this case it’s easier to do things differently. Let's finddt=d(-x 2) = -2xdx. Note that the expression xdx is a factor of the integrand of the desired integral. Let us express it from the resulting equalityxdx= - ½dt. Then
In earlier material, the issue of finding the derivative was considered and its various applications were shown: calculating the slope 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. The 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 based on speed is the physical meaning of an antiderivative.
The resulting function $s(t)$ is called the antiderivative of the function $v(t)$. Quite an interesting and unusual name, isn’t it. It contains a great meaning that explains the essence of 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. It is worth paying special attention to the notation of the antiderivative $F(x)$. First of all, $F$ is capital. Antiderivatives are indicated 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 with a speed equal to zero remains in place, that is, the path it has traveled is 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 antiderivative functions given to us. 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 the 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, we 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.
| 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 is known, many functions have a more complex form than those indicated in the table of antiderivatives, and can be 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 of $x^3$, $\sin x$ 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 |
| $(ax+b)^n, n\ne1, a\ne0$ | $\displaystyle \frac((ax+b)^n)(a(n+1)) +C$ |
| $\displaystyle \frac(1)(ax+b), a\ne0$ | $\displaystyle \frac(1)(a)\ln|ax+b|+C$ |
| $e^(ax+b), a\ne0$ | $\displaystyle \frac(1)(a) e^(ax+b)+C$ |
| $\sin(ax+b), a\ne0$ | $\displaystyle -\frac(1)(a)\cos(ax+b)+C$ |
| $\cos(ax+b), a\ne0$ | $\displaystyle \frac(1)(a)\sin(ax+b)+C$ |
Example 5. Find antiderivatives for:
a) $\displaystyle 4x^3+10x^7$;
b) $\displaystyle \frac(6)(x^5) -\frac(2)(x)$;
c) $\displaystyle 5\cos x+\sin(3x+15)$;
d) $\displaystyle \sqrt(x)-2\sqrt(x)$.
a) $4\frac (x^(3+1))(3+1)+10\frac(x^(7+1))(7+1)+C=x^4+\frac(5)( 4) x^8+C$;
b) $-\frac(3)(2x^4) -2\ln|x|+C$;
c) $5 \sin x - \frac(1)(3)\cos(3x + 15) + C$;
d) $\frac(2)(3)x\sqrt(x) - \frac(3)(2) x\sqrt(x) + C$.
Basic formulas and methods of integration. The rule for integrating a sum or difference. Moving the constant outside the integral sign. Variable replacement method. Formula for integration by parts. An example of solving a problem.
The four main methods of integration are listed below.
1)
The rule for integrating a sum or difference.
.
Here and below u, v, w are functions of the integration variable x.
2)
Moving the constant outside the integral sign.
Let c be a constant independent of x. Then it can be taken out of the integral sign.
3)
Variable replacement method.
Let's consider the indefinite integral.
If we can find such a function φ (x) from x, so
,
then, by replacing the variable t = φ(x) , we have
.
4)
Formula for integration by parts.
,
where u and v are functions of the integration variable.
The ultimate goal of calculating indefinite integrals is, through transformations, to reduce a given integral to the simplest integrals, which are called tabular integrals. Table integrals are expressed through elementary functions using known formulas.
See Table of Integrals >>>
Example
Calculate indefinite integral
Solution
We note that the integrand is the sum and difference of three terms:
, And .
Applying the method 1
.
Next, we note that the integrands of the new integrals are multiplied by constants 5, 4,
And 2
, respectively. Applying the method 2
.
In the table of integrals we find the formula
.
Assuming n = 2
, we find the first integral.
Let us rewrite the second integral in the form
.
We notice that . Then
Let's use the third method. We change the variable t = φ (x) = log x.
.
In the table of integrals we find the formula
Since the variable of integration can be denoted by any letter, then
Let us rewrite the third integral in the form
.
We apply the formula of integration by parts.
Let's put it.
Then
;
;
;
;
.
On this page you will find:
1. Actually, the table of antiderivatives - it can be downloaded in PDF format and printed;
2. Video on how to use this table;
3. A bunch of examples of calculating the antiderivative from various textbooks and tests.
In the video itself, we will analyze many problems where you need to calculate antiderivatives of functions, often quite complex, but most importantly, they are not power functions. All functions summarized in the table proposed above must be known by heart, like derivatives. Without them, further study of integrals and their application to solve practical problems is impossible.
Today we continue to study primitives and move on to a slightly more complex topic. If last time we 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 a completely random function and try to find its derivative, then with a very high probability we will succeed, but the antiderivative will almost never be calculated in this case. But there is good news: there is a fairly large class of functions called elementary functions, the antiderivatives of which are very easy to calculate. And all the other more complex structures that are given on all kinds of tests, independent tests and exams, in fact, are made up of these elementary functions through addition, subtraction and other simple actions. The prototypes of such functions have long been calculated and compiled into special tables. It is these functions and tables that we will work with today.
But we will begin, as always, with a repetition: let’s remember what an antiderivative is, why there are infinitely many of them, and how to determine their general appearance. To do this, I picked up two simple problems.
Solving easy examples
Example #1
Let us immediately note that $\frac(\text( )\!\!\pi\!\!\text( ))(6)$ and in general the presence of $\text( )\!\!\pi\!\!\ text( )$ immediately hints to us that the required antiderivative of the function is related to trigonometry. And, indeed, if we look at the table, we will find that $\frac(1)(1+((x)^(2)))$ is nothing more than $\text(arctg)x$. So let's write it down:
In order to find, you need to write down the following:
\[\frac(\pi )(6)=\text(arctg)\sqrt(3)+C\]
\[\frac(\text( )\!\!\pi\!\!\text( ))(6)=\frac(\text( )\!\!\pi\!\!\text( )) (3)+C\]
Example No. 2
We are also talking about trigonometric functions here. If we look at the table, then, indeed, this is what happens:
We need to find among the entire set of antiderivatives the one that passes through the indicated point:
\[\text( )\!\!\pi\!\!\text( )=\arcsin \frac(1)(2)+C\]
\[\text( )\!\!\pi\!\!\text( )=\frac(\text( )\!\!\pi\!\!\text( ))(6)+C\]
Let's finally write it down:
It's that simple. The only problem is that in order to calculate antiderivatives of simple functions, you need to learn a table of antiderivatives. However, after studying the derivative table for you, I think this will not be a problem.
Solving problems containing an exponential function
To begin with, let's write the following formulas:
\[((e)^(x))\to ((e)^(x))\]
\[((a)^(x))\to \frac(((a)^(x)))(\ln a)\]
Let's see how this all works in practice.
Example #1
If we look at the contents of the brackets, we will notice that in the table of antiderivatives there is no such expression for $((e)^(x))$ to be in a square, so this square must be expanded. To do this, we use the abbreviated multiplication formulas:
Let's find the antiderivative for each of the terms:
\[((e)^(2x))=((\left(((e)^(2)) \right))^(x))\to \frac(((\left(((e)^ (2)) \right))^(x)))(\ln ((e)^(2)))=\frac(((e)^(2x)))(2)\]
\[((e)^(-2x))=((\left(((e)^(-2)) \right))^(x))\to \frac(((\left(((e )^(-2)) \right))^(x)))(\ln ((e)^(-2)))=\frac(1)(-2((e)^(2x))) \]
Now let’s collect all the terms into a single expression and get the general antiderivative:
Example No. 2
This time the degree is larger, so the abbreviated multiplication formula will be quite complex. So let's open the brackets:
Now let’s try to take the antiderivative of our formula from this construction:
As you can see, there is nothing complicated or supernatural in the antiderivatives of the exponential function. All of them are calculated through tables, but attentive students will probably notice that the antiderivative $((e)^(2x))$ is much closer to simply $((e)^(x))$ than to $((a)^(x ))$. So, maybe there is some more special rule that allows, knowing the antiderivative $((e)^(x))$, to find $((e)^(2x))$? Yes, such a rule exists. And, moreover, it is an integral part of working with the table of antiderivatives. We will now analyze it using the same expressions that we just worked with as an example.
Rules for working with the table of antiderivatives
Let's write our function again:
In the previous case, we used the following formula to solve:
\[((a)^(x))\to \frac(((a)^(x)))(\operatorname(lna))\]
But now let’s do it a little differently: let’s remember on what basis $((e)^(x))\to ((e)^(x))$. As I already said, because the derivative $((e)^(x))$ is nothing more than $((e)^(x))$, therefore its antiderivative will be equal to the same $((e) ^(x))$. But the problem is that we have $((e)^(2x))$ and $((e)^(-2x))$. Now let's try to find the derivative of $((e)^(2x))$:
\[((\left(((e)^(2x)) \right))^(\prime ))=((e)^(2x))\cdot ((\left(2x \right))^( \prime ))=2\cdot ((e)^(2x))\]
Let's rewrite our construction again:
\[((\left(((e)^(2x)) \right))^(\prime ))=2\cdot ((e)^(2x))\]
\[((e)^(2x))=((\left(\frac(((e)^(2x)))(2) \right))^(\prime ))\]
This means that when we find the antiderivative $((e)^(2x))$ we get the following:
\[((e)^(2x))\to \frac(((e)^(2x)))(2)\]
As you can see, we got the same result as before, but we did not use the formula to find $((a)^(x))$. Now this may seem stupid: why complicate the calculations when there is a standard formula? However, in slightly more complex expressions you will find that this technique is very effective, i.e. using derivatives to find antiderivatives.
As a warm-up, let's find the antiderivative of $((e)^(2x))$ in a similar way:
\[((\left(((e)^(-2x)) \right))^(\prime ))=((e)^(-2x))\cdot \left(-2 \right)\]
\[((e)^(-2x))=((\left(\frac(((e)^(-2x)))(-2) \right))^(\prime ))\]
When calculating, our construction will be written as follows:
\[((e)^(-2x))\to -\frac(((e)^(-2x)))(2)\]
\[((e)^(-2x))\to -\frac(1)(2\cdot ((e)^(2x)))\]
We got exactly the same result, but took a different path. It is this path, which now seems a little more complicated to us, that in the future will turn out to be more effective for calculating more complex antiderivatives and using tables.
Note! This is a very important point: antiderivatives, like derivatives, can be counted in many different ways. However, if all calculations and calculations are equal, then the answer will be the same. We have just seen this 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 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.
More complex problems for solving antiderivative exponential functions
Task No. 1
Let's note the following:
\[((2)^(x))\cdot ((5)^(x))=((\left(2\cdot 5 \right))^(x))=((10)^(x) )\]
To find the antiderivative of this expression, simply use the standard formula - $((a)^(x))\to \frac(((a)^(x)))(\ln a)$.
In our case, the antiderivative will be like this:
Of course, compared to the design we just solved, this one looks simpler.
Problem No. 2
Again, it's easy to see that this function can easily be divided into two separate terms - two separate fractions. Let's rewrite:
It remains to find the antiderivative of each of these terms using the formula described above:
Despite the apparent greater complexity of exponential functions compared to power functions, the overall volume of calculations and calculations turned out to be much simpler.
Of course, for knowledgeable students, what we have just discussed (especially against the backdrop of what we have discussed before) may seem like elementary expressions. However, when choosing these two problems for today's video lesson, I did not set myself the goal of telling you another complex and sophisticated technique - all I wanted to show you is that you should not be afraid to use standard algebra techniques to transform original functions.
Using a "secret" technique
In conclusion, I would like to look at another interesting technique, which, on the one hand, goes beyond what we mainly discussed today, but, on the other hand, it is, firstly, not at all complicated, i.e. Even beginner students can master it, and, secondly, it is quite often found in all kinds of tests and independent work, i.e. knowledge of it will be very useful in addition to knowledge of the table of antiderivatives.
Task No. 1
Obviously, we have something very similar to a power function. What should we do in this case? Let's think about it: $x-5$ is not that much different from $x$ - they just added $-5$. Let's write it like this:
\[((x)^(4))\to \frac(((x)^(5)))(5)\]
\[((\left(\frac(((x)^(5)))(5) \right))^(\prime ))=\frac(5\cdot ((x)^(4))) (5)=((x)^(4))\]
Let's try to find the derivative of $((\left(x-5 \right))^(5))$:
\[((\left(((\left(x-5 \right))^(5)) \right))^(\prime ))=5\cdot ((\left(x-5 \right)) ^(4))\cdot ((\left(x-5 \right))^(\prime ))=5\cdot ((\left(x-5 \right))^(4))\]
This implies:
\[((\left(x-5 \right))^(4))=((\left(\frac(((\left(x-5 \right))^(5)))(5) \ right))^(\prime ))\]
There is no such value in the table, so we have now derived this formula ourselves using the standard antiderivative formula for a power function. Let's write the answer like this:
Problem No. 2
Many students who look at the first solution may think that everything is very simple: just replace $x$ in the power function with a linear expression, and everything will fall into place. Unfortunately, everything is not so simple, and now we will see this.
By analogy with the first expression, we write the following:
\[((x)^(9))\to \frac(((x)^(10)))(10)\]
\[((\left(((\left(4-3x \right))^(10)) \right))^(\prime ))=10\cdot ((\left(4-3x \right)) ^(9))\cdot ((\left(4-3x \right))^(\prime ))=\]
\[=10\cdot ((\left(4-3x \right))^(9))\cdot \left(-3 \right)=-30\cdot ((\left(4-3x \right)) ^(9))\]
Returning to our derivative, we can write:
\[((\left(((\left(4-3x \right))^(10)) \right))^(\prime ))=-30\cdot ((\left(4-3x \right) )^(9))\]
\[((\left(4-3x \right))^(9))=((\left(\frac(((\left(4-3x \right))^(10)))(-30) \right))^(\prime ))\]
This immediately follows:
Nuances of the solution
Please note: if nothing essentially changed last time, then in the second case, instead of $-10$, $-30$ appeared. What is the difference between $-10$ and $-30$? Obviously, by a factor of $-3$. Question: where did it come from? If you look closely, you can see that it was taken as a result of calculating the derivative of a complex function - the coefficient that stood at $x$ appears in the antiderivative below. This is a very important rule, which I initially did not plan to discuss at all in today’s video lesson, but without it the presentation of tabular antiderivatives would be incomplete.
So let's do it again. Let there be our main power function:
\[((x)^(n))\to \frac(((x)^(n+1)))(n+1)\]
Now, instead of $x$, let's substitute the expression $kx+b$. What will happen then? We need to find the following:
\[((\left(kx+b \right))^(n))\to \frac(((\left(kx+b \right))^(n+1)))(\left(n+ 1\right)\cdot k)\]
On what basis do we claim this? Very simple. Let's find the derivative of the construction written above:
\[((\left(\frac(((\left(kx+b \right))^(n+1)))(\left(n+1 \right)\cdot k) \right))^( \prime ))=\frac(1)(\left(n+1 \right)\cdot k)\cdot \left(n+1 \right)\cdot ((\left(kx+b \right))^ (n))\cdot k=((\left(kx+b \right))^(n))\]
This is the same expression that originally existed. Thus, this formula is also correct, and it can be used to supplement the table of antiderivatives, or it is better to simply memorize the entire table.
Conclusions from the “secret: technique:
- Both functions that we just looked at can, in fact, be reduced to the antiderivatives indicated in the table by expanding the degrees, but if we can more or less somehow cope with the fourth degree, then I wouldn’t even consider the ninth degree dared to reveal.
- If we were to expand the degrees, we would end up with such a volume of calculations that a simple task would take us an inappropriately large amount of time.
- That is why such problems, which contain linear expressions, do not need to be solved “headlong”. As soon as you come across an antiderivative that differs from the one in the table only by the presence of the expression $kx+b$ inside, immediately remember the formula written above, substitute it into your table antiderivative, and everything will turn out much faster and easier.
Naturally, due to the complexity and seriousness of this technique, we will return to its consideration many times in future video lessons, but that’s all for today. I hope this lesson will really help those students who want to understand antiderivatives and integration.