Integral dx root of 1 4x 2. Solving the integral online

Finding the indefinite integral is a very common problem in higher mathematics and other technical branches of science. Even the solution of the simplest physical problems is often not complete without the calculation of several simple integrals. Therefore, from school age, we are taught techniques and methods for solving integrals, numerous tables with integrals of the simplest functions are given. However, over time, all this is safely forgotten, either we do not have enough time for calculations or we need to find a solution to the indefinite integral from a very complex function. To solve these problems, our service will be indispensable for you, which allows you to accurately find the indefinite integral online.

Solve the indefinite integral

Online service on website allows you to find online integral solution fast, free and high quality. You can replace the search in the tables of the required integral with our service, where by quickly entering the desired functions, you will get the solution of the indefinite integral in a tabular version. Not all mathematical sites are able to calculate indefinite integrals of functions online quickly and efficiently, especially if you need to find indefinite integral from a complex function or such functions that are not included in the general course of higher mathematics. Website website will help solve the integral online and cope with the task. Using the online solution of the integral on the site site, you will always get the exact answer.

Even if you want to calculate the integral yourself, thanks to our service it will be easy for you to check your answer, find a mistake or typo, or make sure that the task is completed flawlessly. If you are solving a problem and you need to calculate the indefinite integral as an auxiliary action, then why waste time on these actions, which you may have already performed a thousand times? Moreover, additional calculations of the integral can be the cause of a typo or a small error, which subsequently led to an incorrect answer. Just use our services and find indefinite integral online without any effort. For practical tasks of finding integral functions online this server is very helpful. You need to enter a given function, get online indefinite integral solution and compare the answer with your solution.

An irrational function of a variable is a function that is formed from a variable and arbitrary constants using a finite number of operations of addition, subtraction, multiplication (raising to an integer power), division, and extraction of roots. An irrational function differs from a rational one in that an irrational function contains root extraction operations.

There are three main types of irrational functions whose indefinite integrals can be reduced to integrals of rational functions. These are integrals containing roots of arbitrary integer powers from a linear-fractional function (the roots can be of different degrees, but from the same linear-fractional function); integrals of the differential binomial and integrals with a square root of a square trinomial.

Important note. Roots are meaningful!

When calculating integrals containing roots, one often encounters expressions of the form , where is some function of the integration variable . In doing so, it should be borne in mind that . That is, for t > 0 , |t| = t. At t< 0 , |t| = - t . Therefore, when calculating such integrals, it is necessary to separately consider the cases t > 0 and t< 0 . This can be done by writing signs or where necessary. Assuming that the upper sign refers to the case t > 0 , and the lower one - to the case t< 0 . With further transformation, these signs, as a rule, are mutually reduced.

A second approach is also possible, in which the integrand and the result of integration can be considered as complex functions of complex variables. Then you can not follow the signs in the radical expressions. This approach is applicable if the integrand is analytic, that is, a differentiable function of a complex variable. In this case, both the integrand and its integral are multivalued functions. Therefore, after integration, when substituting numerical values, it is necessary to select a single-valued branch (Riemann surface) of the integrand, and for it select the corresponding branch of the integration result.

Fractional linear irrationality

These are integrals with roots of the same linear-fractional function:
,
where R is a rational function, are rational numbers, m 1 , n 1 , ..., m s , n s are integers, α, β, γ, δ are real numbers.
Such integrals are reduced to the integral of a rational function by substitution:
, where n is the common denominator of numbers r 1 , ..., r s .

The roots may not necessarily be from a linear-fractional function, but also from a linear one (γ = 0 , δ = 1), or from the integration variable x (α = 1 , β = 0 , γ = 0 , δ = 1).

Here are examples of such integrals:
, .

Integrals from differential binomials

Integrals from differential binomials have the form:
,
where m, n, p are rational numbers, a, b are real numbers.
Such integrals reduce to integrals of rational functions in three cases.

1) If p is an integer. Substitution x = t N , where N is the common denominator of the fractions m and n .
2) If is an integer. Substitution a x n + b = t M , where M is the denominator of p .
3) If is an integer. Substitution a + b x - n = t M , where M is the denominator of p .

In other cases, such integrals are not expressed in terms of elementary functions.

Sometimes such integrals can be simplified using reduction formulas:
;
.

Integrals Containing the Square Root of a Square Trinomial

Such integrals have the form:
,
where R is a rational function. For each such integral, there are several methods for solving it.
1) With the help of transformations lead to simpler integrals.
2) Apply trigonometric or hyperbolic substitutions.
3) Apply Euler substitutions.

Let's consider these methods in more detail.

1) Transformation of the integrand

Applying the formula and performing algebraic transformations, we bring the integrand to the form:
,
where φ(x), ω(x) are rational functions.

I type

Integral of the form:
,
where P n (x) is a polynomial of degree n.

Such integrals are found by the method of indefinite coefficients, using the identity:

.
Differentiating this equation and equating the left and right sides, we find the coefficients A i .

II type

Integral of the form:
,
where P m (x) is a polynomial of degree m.

Substitution t = (x - α) -1 this integral is reduced to the previous type. If m ≥ n, then the fraction should have an integer part.

III type

Here we make a substitution:
.
Then the integral will take the form:
.
Further, the constants α, β must be chosen such that the coefficients at t in the denominator vanish:
B = 0, B 1 = 0 .
Then the integral decomposes into the sum of integrals of two types:
,
,
which are integrated by substitutions:
u 2 \u003d A 1 t 2 + C 1,
v 2 \u003d A 1 + C 1 t -2.

2) Trigonometric and hyperbolic substitutions

For integrals of the form , a > 0 ,
we have three main substitutions:
;
;
;

For integrals , a > 0 ,
we have the following substitutions:
;
;
;

And, finally, for integrals , a > 0 ,
substitutions are as follows:
;
;
;

3) Euler substitutions

Integrals can also be reduced to integrals of rational functions of one of the three Euler substitutions:
, for a > 0 ;
, for c > 0 ;
, where x 1 is the root of the equation a x 2 + b x + c = 0. If this equation has real roots.

Elliptic integrals

Finally, consider integrals of the form:
,
where R is a rational function, . Such integrals are called elliptic. In general, they are not expressed in terms of elementary functions. However, there are cases when there are relations between the coefficients A, B, C, D, E, in which such integrals are expressed in terms of elementary functions.

The following is an example related to recursive polynomials. The calculation of such integrals is performed using substitutions:
.

Example

Calculate integral:
.

Solution

We make a substitution.

.
Here, for x > 0 (u > 0 ) we take the upper sign ′+ ′. For x< 0 (u< 0 ) - lower ′- ′.

.

Answer

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

A function F(x) differentiable in a given interval X is called antiderivative for the function f(x), or an integral of f(x) if for any x ∈X the equality holds:

F "(x) = f(x). (8.1)

Finding all antiderivatives for a given function is called its integration. The indefinite integral of the function f(x) on a given interval X is the set of all antiderivatives for the function f(x); designation -

If F(x) is some antiderivative for the function f(x), then ∫ f(x)dx = F(x) + C, (8.2)

where C is an arbitrary constant.

Table of integrals

Directly from the definition we obtain the main properties of the indefinite integral and the list of table integrals:

1) d∫f(x)dx=f(x)

2)∫df(x)=f(x)+C

3) ∫af(x)dx=a∫f(x)dx (a=const)

4) ∫(f(x)+g(x))dx = ∫f(x)dx+∫g(x)dx

List of table integrals

1. ∫x m dx = x m+1 /(m + 1) +C; (m ≠ -1)

3.∫a x dx = a x / ln a + C (a>0, a ≠1)

4.∫e x dx = e x + C

5.∫sin x dx = cosx + C

6.∫cos x dx = - sin x + C

7. = arctg x + C

8.=arcsin x + C

10.=-ctg x + C

Variable substitution

To integrate many functions, the method of changing a variable is used or substitutions, allowing to bring integrals to a tabular form.

If the function f(z) is continuous on [α,β], the function z =g(x) has a continuous derivative and α ≤ g(x) ≤ β, then

∫ f(g(x)) g " (x) dx = ∫f(z)dz, (8.3)

moreover, after integration on the right side, one should make a substitution z=g(x).

To prove it, it suffices to write the original integral in the form:

∫ f(g(x)) g " (x) dx = ∫ f(g(x)) dg(x).

For example:

Method of integration by parts

Let u = f(x) and v = g(x) be functions having continuous . Then, according to the works,

d(uv))= udv + vdu or udv = d(uv) - vdu.

For the expression d(uv), the antiderivative will obviously be uv, so the formula takes place:

∫ udv = uv - ∫ vdu (8.4.)

This formula expresses the rule integration by parts. It brings the integration of the expression udv=uv"dx to the integration of the expression vdu=vu"dx.

Let, for example, it is required to find ∫xcosx dx. Let u = x, dv = cosxdx, so du=dx, v=sinx. Then

∫xcosxdx = ∫x d(sin x) = x sin x - ∫sin x dx = x sin x + cosx + C.

The rule of integration by parts has a more limited scope than the change of variable. But there are whole classes of integrals, for example,

∫x k ln m xdx, ∫x k sinbxdx, ∫ x k cosbxdx, ∫x k e ax and others, which are calculated exactly using integration by parts.

Definite integral

The concept of a definite integral is introduced as follows. Let a function f(x) be defined on an interval. Let us split the segment [a,b] into n parts by points a= x 0< x 1 <...< x n = b. Из каждого интервала (x i-1 , x i) возьмем произвольную точку ξ i и составим сумму f(ξ i) Δx i где
Δ x i \u003d x i - x i-1. The sum of the form f(ξ i)Δ x i is called integral sum, and its limit at λ = maxΔx i → 0, if it exists and is finite, is called definite integral functions f(x) of a before b and is denoted:

F(ξ i)Δx i (8.5).

The function f(x) in this case is called integrable on a segment, the numbers a and b are called lower and upper limit of the integral.

The following properties hold for a definite integral:

4), (k = const, k∈R);

5)

6)

7) f(ξ)(b-a) (ξ∈).

The last property is called mean value theorem.

Let f(x) be continuous on . Then on this segment there exists an indefinite integral

∫f(x)dx = F(x) + C

and takes place Newton-Leibniz formula, which connects the definite integral with the indefinite one:

F(b) - F(a). (8.6)

Geometric interpretation: the definite integral is the area of ​​a curvilinear trapezoid bounded from above by the curve y=f(x), the straight lines x = a and x = b and the axis segment Ox.

Improper integrals

Integrals with infinite limits and integrals of discontinuous (unbounded) functions are called improper. Improper integrals of the first kind - these are integrals over an infinite interval, defined as follows:

(8.7)

If this limit exists and is finite, then it is called convergent improper integral of f(x) on the interval [а,+ ∞), and the function f(x) is called integrable on an infinite interval[a,+ ∞). Otherwise, the integral is said to be does not exist or diverges.

The improper integrals on the intervals (-∞,b] and (-∞, + ∞) are defined similarly:

Let us define the concept of an integral of an unbounded function. If f(x) is continuous for all values x segment , except for the point c, at which f(x) has an infinite discontinuity, then improper integral of the second kind of f(x) ranging from a to b called the sum:

if these limits exist and are finite. Designation:

Examples of calculating integrals

Example 3.30. Calculate ∫dx/(x+2).

Solution. Denote t = x+2, then dx = dt, ∫dx/(x+2) = ∫dt/t = ln|t| + C = ln|x+2| +C.

Example 3.31. Find ∫ tgxdx.

Solution.∫tgxdx = ∫sinx/cosxdx = - ∫dcosx/cosx. Let t=cosx, then ∫ tgxdx = -∫ dt/t = - ln|t| + C = -ln|cosx|+C.

Solution.

Example3.33. Find .

Solution. = .

Example3.34 . Find ∫arctgxdx.

Solution. We integrate by parts. Denote u=arctgx, dv=dx. Then du = dx/(x 2 +1), v=x, whence ∫arctgxdx = xarctgx - ∫ xdx/(x 2 +1) = xarctgx + 1/2 ln(x 2 +1) +C; because
∫xdx/(x 2 +1) = 1/2 ∫d(x 2 +1)/(x 2 +1) = 1/2 ln(x 2 +1) +C.

Example3.35 . Calculate ∫lnxdx.

Solution. Applying the integration-by-parts formula, we get:
u=lnx, dv=dx, du=1/x dx, v=x. Then ∫lnxdx = xlnx - ∫x 1/x dx =
= xlnx - ∫dx + C= xlnx - x + C.

Example3.36 . Calculate ∫e x sinxdx.

Solution. Denote u = e x , dv = sinxdx, then du = e x dx, v =∫sinxdx= - cosx → ∫ e x sinxdx = - e x cosx + ∫ e x cosxdx. The integral ∫e x cosxdx is also integrable by parts: u = e x , dv = cosxdx, du=e x dx, v=sinx. We have:
∫ e x cosxdx = e x sinx - ∫ e x sinxdx. We got the relation ∫e x sinxdx = - e x cosx + e x sinx - ∫ e x sinxdx, whence 2∫e x sinx dx = - e x cosx + e x sinx + C.

Example 3.37. Calculate J = ∫cos(lnx)dx/x.

Solution. Since dx/x = dlnx, then J= ∫cos(lnx)d(lnx). Replacing lnx through t, we arrive at the table integral J = ∫ costdt = sint + C = sin(lnx) + C.

Example 3.38 . Calculate J = .

Solution. Taking into account that = d(lnx), we make the substitution lnx = t. Then J = .

Example 3.39 . Calculate the integral J = .

Solution. We have: . Therefore =
=
=. entered as sqrt(tan(x/2)).

And if you click on Show steps in the upper right corner in the result window, you will get a detailed solution.