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 out the constant factor and/or representing the integrand as a sum of functions - decomposition of the integrand into terms).
Example 1. For example, to find(dx/x 4) you can directly use the table integral 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 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 
, 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
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 special case formulas (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 sinx functions 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 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: does not exist 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 = ? (30)
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 power function, sine, exponential and constant 1. Let’s not 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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Definition 1
The antiderivative $F(x)$ for the function $y=f(x)$ on the segment $$ is a function that is differentiable at each point of this segment and the following equality holds for its derivative:
Definition 2
The set of all antiderivatives given function$y=f(x)$ defined on a certain segment is called the indefinite integral of a given function $y=f(x)$. The indefinite integral is denoted by the symbol $\int f(x)dx $.
From the table of derivatives and Definition 2 we obtain the table of basic integrals.
Example 1
Check the validity of formula 7 from the table of integrals:
\[\int tgxdx =-\ln |\cos x|+C,\, \, C=const.\]
Let's differentiate the right-hand side: $-\ln |\cos x|+C$.
\[\left(-\ln |\cos x|+C\right)"=-\frac(1)(\cos x) \cdot (-\sin x)=\frac(\sin x)(\cos x) =tgx\]
Example 2
Check the validity of formula 8 from the table of integrals:
\[\int ctgxdx =\ln |\sin x|+C,\, \, C=const.\]
Let's differentiate the right-hand side: $\ln |\sin x|+C$.
\[\left(\ln |\sin x|\right)"=\frac(1)(\sin x) \cdot \cos x=ctgx\]
The derivative turned out to be equal to the integrand. Therefore, the formula is correct.
Example 3
Check the validity of formula 11" from the table of integrals:
\[\int \frac(dx)(a^(2) +x^(2) ) =\frac(1)(a) arctg\frac(x)(a) +C,\, \, C=const .\]
Let's differentiate the right-hand side: $\frac(1)(a) arctg\frac(x)(a) +C$.
\[\left(\frac(1)(a) arctg\frac(x)(a) +C\right)"=\frac(1)(a) \cdot \frac(1)(1+\left( \frac(x)(a) \right)^(2) ) \cdot \frac(1)(a) =\frac(1)(a^(2) ) \cdot \frac(a^(2) ) (a^(2) +x^(2) ) \]
The derivative turned out to be equal to the integrand. Therefore, the formula is correct.
Example 4
Check the validity of formula 12 from the table of integrals:
\[\int \frac(dx)(a^(2) -x^(2) ) =\frac(1)(2a) \ln \left|\frac(a+x)(a-x) \right|+ C,\, \, C=const.\]
Let's differentiate the right-hand side: $\frac(1)(2a) \ln \left|\frac(a+x)(a-x) \right|+C$.
$\left(\frac(1)(2a) \ln \left|\frac(a+x)(a-x) \right|+C\right)"=\frac(1)(2a) \cdot \frac( 1)(\frac(a+x)(a-x) ) \cdot \left(\frac(a+x)(a-x) \right)"=\frac(1)(2a) \cdot \frac(a-x)( a+x) \cdot \frac(a-x+a+x)((a-x)^(2) ) =\frac(1)(2a) \cdot \frac(a-x)(a+x) \cdot \ frac(2a)((a-x)^(2) ) =\frac(1)(a^(2) -x^(2) ) $The derivative turned out to be equal to the integrand. Therefore, the formula is correct.
Example 5
Check the validity of formula 13" from the table of integrals:
\[\int \frac(dx)(\sqrt(a^(2) -x^(2) ) ) =\arcsin \frac(x)(a) +C,\, \, C=const.\]
Let's differentiate the right-hand side: $\arcsin \frac(x)(a) +C$.
\[\left(\arcsin \frac(x)(a) +C\right)"=\frac(1)(\sqrt(1-\left(\frac(x)(a) \right)^(2 ) ) ) \cdot \frac(1)(a) =\frac(a)(\sqrt(a^(2) -x^(2) ) ) \cdot \frac(1)(a) =\frac( 1)(\sqrt(a^(2) -x^(2) ) ) \]
The derivative turned out to be equal to the integrand. Therefore, the formula is correct.
Example 6
Check the validity of formula 14 from the table of integrals:
\[\int \frac(dx)(\sqrt(x^(2) \pm a^(2) ) =\ln |x+\sqrt(x^(2) \pm a^(2) ) |+ C,\, \, C=const.\]
Let's differentiate the right-hand side: $\ln |x+\sqrt(x^(2) \pm a^(2) ) |+C$.
\[\left(\ln |x+\sqrt(x^(2) \pm a^(2) ) |+C\right)"=\frac(1)(x+\sqrt(x^(2) \pm a^(2) ) \cdot \left(x+\sqrt(x^(2) \pm a^(2) ) \right)"=\frac(1)(x+\sqrt(x^(2) \ pm a^(2) ) \cdot \left(1+\frac(1)(2\sqrt(x^(2) \pm a^(2) ) ) \cdot 2x\right)=\] \[ =\frac(1)(x+\sqrt(x^(2) \pm a^(2) ) \cdot \frac(\sqrt(x^(2) \pm a^(2) ) +x)( \sqrt(x^(2) \pm a^(2) ) =\frac(1)(\sqrt(x^(2) \pm a^(2) ) ) \]
The derivative turned out to be equal to the integrand. Therefore, the formula is correct.
Example 7
Find the integral:
\[\int \left(\cos (3x+2)+5x\right) dx.\]
Let's use the sum integral theorem:
\[\int \left(\cos (3x+2)+5x\right) dx=\int \cos (3x+2)dx +\int 5xdx .\]
Let us use the theorem about placing a constant factor outside the integral sign:
\[\int \cos (3x+2)dx +\int 5xdx =\int \cos (3x+2)dx +5\int xdx .\]
According to the table of integrals:
\[\int \cos x dx=\sin x+C;\] \[\int xdx =\frac(x^(2) )(2) +C.\]
When calculating the first integral, we use rule 3:
\[\int \cos (3x+2) dx=\frac(1)(3) \sin (3x+2)+C_(1) .\]
Hence,
\[\int \left(\cos (3x+2)+5x\right) dx=\frac(1)(3) \sin (3x+2)+C_(1) +\frac(5x^(2) )(2) +C_(2) =\frac(1)(3) \sin (3x+2)+\frac(5x^(2) )(2) +C,\, \, C=C_(1 ) +C_(2) \]
Integration is one of the main operations in mathematical analysis. Tables of known antiderivatives can be useful, but now, after the advent of computer algebra systems, they are losing their significance. Below is a list of the most common primitives.
Table of basic integrals
Another, compact option
Table of integrals of trigonometric functions
From rational functions
From irrational functions
Integrals of transcendental functions
"C" is an arbitrary integration constant, which is determined if the value of the integral at any point is known. Each function has an infinite number of antiderivatives.
Most schoolchildren and students have problems calculating integrals. This page contains integral tables from trigonometric, rational, irrational and transcendental functions that will help in the solution. It will also help you derivative table.
