Integration of the simplest irrational functions. Integrating Complex Fractions
Let us consider integrals with fractional roots linear function:
(1)
,
where R is the rational function of its arguments. That is, a function composed of its arguments and arbitrary constants using finite number operations of addition (subtraction), multiplication and division (raising to an integer power).
Examples of considered integrals with linear-fractional irrationality
Let us give examples of integrals with roots of the form (1) .
Example 1
Although here the integral sign includes roots of various degrees, the integrand expression can be transformed as follows:
;
;
.
Thus, the integrand is made up of the integration variable x and the root of the linear function using a finite number of subtraction, division, and multiplication operations. Therefore it is rational function from x and and belongs to the type in question (1)
with constant values n = 6
,
α = β = δ = 1
,
γ = 0
:
.
Example 2
Here we do the conversion:
.
This shows that the integrand is a rational function of x and . Therefore it belongs to the type in question.
General example of fractional linear irrationality
In a more general case, the integrand can include any finite number of roots of the same linear fractional function:
(2)
,
where R is the rational function of its arguments,
- rational numbers,
m 1, n 1, ..., m s, n s- integers.
Indeed, let n be the common denominator of the numbers r 1, ..., r s. Then they can be represented as:
,
where k 1 , k 2 , ..., k s- integers. Then everyone included in (2)
roots are powers of:
,
,
. . . . .
.
That is, the entire integrand (2)
made up of x and the root using a finite number of addition, multiplication, and division operations. Therefore it is a rational function of x and :
.
Root integration method
Integral with fractional linear irrationality
(1)
reduces to the integral of a rational function by substitution
(3)
.
Proof
Extract the nth root from both sides (3)
:
.
Let's transform (3)
:
;
;
.
Finding the derivative:
;
;
.
Differential:
.
Substitute in (1)
:
.
From this it is clear that integrand is composed of constants and an integration variable t using a finite number of addition (subtraction), multiplication (raising to an integer power) and division operations. Therefore, the integrand is a rational function of the integration variable. Thus, the calculation of the integral was reduced to the integration of a rational function. Q.E.D.
Example of linear irrationality integration
Find the integral:
Solution
Since the integral includes roots of the same (fractional) linear function x + 1 , and the integrand is formed using the operations of subtraction and division, then this integral belongs to the type under consideration.
Let us transform the integrand so that it includes roots of the same degree:
;
;
.
Making a substitution
x+ 1 = t 6.
Let's take the differential:
d (x + 1) = dx = (
t 6 )′
dt = 6 t 5 dt.
Let's substitute:
x = t 6 - 1
;
;
;
.
We select the whole part of the fraction, noting that
t 6 - 1 =
(t - 1)(t 5 + t 4 + t 3 + t 2 + t + 1).
Then
.
Answer
,
Where .
Example of integration of fractional-linear irrationality
Find the integral
Solution
Let us select the root of the linear fractional function:
.
Then
.
Making a substitution
.
Take the differential
.
Finding the derivative
.
Then
.
Next we notice that
.
Substitute into the integrand
.
Answer
Used literature:
N.M. Gunter, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003.
We continue to consider integrals of fractions and roots. Not all of them are super complex, it’s just that for one reason or another the examples were a little “off topic” in other articles.
Example 9
Find the indefinite integral
In the denominator under the root there is a quadratic trinomial plus an “appendage” in the form of an “X” outside the root. An integral of this type can be solved using a standard substitution.
.
The replacement here is simple:
Let's look at life after replacement:
(1) After substitution, we reduce the terms under the root to a common denominator.
(2) We take it out from under the root.
(3) The numerator and denominator are reduced by . At the same time, under the root, we rearranged the terms into convenient order. With some experience, steps (1), (2) can be skipped by performing the commented actions orally.
(4) The resulting integral, as you remember, is solved complete square extraction method. Select a complete square.
(5) By integration we obtain an ordinary “long” logarithm.
(6) We carry out the reverse replacement. If initially , then back: .
(7) Final action aimed at styling the result: under the root we again bring the terms to a common denominator and take them out from under the root.
Example 10
Find the indefinite integral
.
This is an example for independent decision. Here a constant is added to the lone “X”, and the replacement is almost the same:
.
The only thing that is needed is to additionally express the “x” from the replacement being carried out:
.
Complete solution and the answer at the end of the lesson.
Sometimes in such an integral there may be a quadratic binomial under the root, this does not change the method of solution, it will be even simpler. Feel the difference:
Example 11
Find the indefinite integral
Example 12
Find the indefinite integral
Brief solutions and answers at the end of the lesson. It should be noted that Example 11 is exactly binomial integral, the solution of which was discussed in class Integrals from irrational functions .
Integral of a polynomial of the 2nd degree indecomposable in the denominator to the power
More rare, but nevertheless found in practical examples type of integral.
Example 13
Find the indefinite integral
The denominator of the integrand contains a quadratic binomial that cannot be factorized. We emphasize that non-factorizability is an essential feature. If the polynomial is factorized, then everything is much clearer, for example:
Let's go back to the example with lucky number 13. This integral is also one of those that can be quite painful if you don’t know how to solve.
The solution starts with an artificial transformation:
I think everyone already understands how to divide the numerator by the denominator term by term.
The resulting integral is taken in parts:
For an integral of the form
Where ( k≥ 2) – natural number, derived recurrent reduction formula:
; is an integral of a degree lower by 1.
What if there is an additional polynomial in the numerator? In this case, the method of indefinite coefficients is used, and the integrand is expanded into a sum of fractions. If you encounter such an integral, look at the textbook - everything is simple there.
Definition 1
The set of all antiderivatives given function$y=f(x)$ defined on a certain segment is called the indefinite integral of a given function $y=f(x)$. Indefinite integral denoted by the symbol $\int f(x)dx $.
Comment
Definition 2 can be written as follows:
\[\int f(x)dx =F(x)+C.\]
Not every irrational function can be expressed as an integral through elementary functions. However, most of these integrals can be reduced using substitutions to integrals of rational functions, which can be expressed in terms of elementary functions.
$\int R\left(x,x^(m/n) ,...,x^(r/s) \right)dx $;
$\int R\left(x,\left(\frac(ax+b)(cx+d) \right)^(m/n) ,...,\left(\frac(ax+b)(cx +d) \right)^(r/s) \right)dx $;
$\int R\left(x,\sqrt(ax^(2) +bx+c) \right)dx $.
I
When finding an integral of the form $\int R\left(x,x^(m/n) ,...,x^(r/s) \right)dx $ it is necessary to perform the following substitution:
With this substitution, each fractional power of the variable $x$ is expressed through an integer power of the variable $t$. As a result, the integrand function is transformed into a rational function of the variable $t$.
Example 1
Perform integration:
\[\int \frac(x^(1/2) dx)(x^(3/4) +1) .\]
Solution:
$k=4$ is the common denominator of the fractions $\frac(1)(2) ,\, \, \frac(3)(4) $.
\ \[\begin(array)(l) (\int \frac(x^(1/2) dx)(x^(3/4) +1) =4\int \frac(t^(2) ) (t^(3) +1) \cdot t^(3) dt =4\int \frac(t^(5) )(t^(3) +1) dt =4\int \left(t^( 2) -\frac(t^(2) )(t^(3) +1) \right)dt =4\int t^(2) dt -4\int \frac(t^(2) )(t ^(3) +1) dt =\frac(4)(3) \cdot t^(3) -) \\ (-\frac(4)(3) \cdot \ln |t^(3) +1 |+C)\end(array)\]
\[\int \frac(x^(1/2) dx)(x^(3/4) +1) =\frac(4)(3) \cdot \left+C\]
II
When finding an integral of the form $\int R\left(x,\left(\frac(ax+b)(cx+d) \right)^(m/n) ,...,\left(\frac(ax+ b)(cx+d) \right)^(r/s) \right)dx $ it is necessary to perform the following substitution:
where $k$ is the common denominator of the fractions $\frac(m)(n) ,...,\frac(r)(s) $.
As a result of this substitution, the integrand is transformed into a rational function of the variable $t$.
Example 2
Perform integration:
\[\int \frac(\sqrt(x+4) )(x) dx .\]
Solution:
Let's make the following substitution:
\ \[\int \frac(\sqrt(x+4) )(x) dx =\int \frac(t^(2) )(t^(2) -4) dt =2\int \left(1 +\frac(4)(t^(2) -4) \right)dt =2\int dt +8\int \frac(dt)(t^(2) -4) =2t+2\ln \left |\frac(t-2)(t+2) \right|+C\]
After making the reverse substitution, we get the final result:
\[\int \frac(\sqrt(x+4) )(x) dx =2\sqrt(x+4) +2\ln \left|\frac(\sqrt(x+4) -2)(\ sqrt(x+4) +2) \right|+C.\]
III
When finding an integral of the form $\int R\left(x,\sqrt(ax^(2) +bx+c) \right)dx $, the so-called Euler substitution is performed (one of three possible substitutions is used).
Euler's first substitution
For the case $a>
Taking the “+” sign in front of $\sqrt(a) $, we get
Example 3
Perform integration:
\[\int \frac(dx)(\sqrt(x^(2) +c) ) .\]
Solution:
Let's make the following substitution (case $a=1>0$):
\[\sqrt(x^(2) +c) =-x+t,\, \, x=\frac(t^(2) -c)(2t) ,\, \, dx=\frac(t ^(2) +c)(2t^(2) ) dt,\, \, \sqrt(x^(2) +c) =-\frac(t^(2) -c)(2t) +t= \frac(t^(2) +c)(2t) .\] \[\int \frac(dx)(\sqrt(x^(2) +c) ) =\int \frac(\frac(t^ (2) +c)(2t^(2) ) dt)(\frac(t^(2) +c)(2t) ) =\int \frac(dt)(t) =\ln |t|+C \]
After making the reverse substitution, we get the final result:
\[\int \frac(dx)(\sqrt(x^(2) +c) ) =\ln |\sqrt(x^(2) +c) +x|+C.\]
Euler's second substitution
For the case $c>0$ it is necessary to perform the following substitution:
Taking the “+” sign in front of $\sqrt(c) $, we get
Example 4
Perform integration:
\[\int \frac((1-\sqrt(1+x+x^(2) ))^(2) )(x^(2) \sqrt(1+x+x^(2) ) ) dx .\]
Solution:
Let's make the following substitution:
\[\sqrt(1+x+x^(2) ) =xt+1.\]
\ \[\sqrt(1+x+x^(2) ) =xt+1=\frac(t^(2) -t+1)(1-t^(2) ) \] \
$\int \frac((1-\sqrt(1+x+x^(2) ))^(2) )(x^(2) \sqrt(1+x+x^(2) ) ) dx = \int \frac((-2t^(2) +t)^(2) (1-t)^(2) (1-t^(2))(2t^(2) -2t+2))( (1-t^(2))^(2) (2t-1)^(2) (t^(2) -t+1)(1-t^(2))^(2) ) dt =\ int \frac(t^(2) )(1-t^(2) ) dt =-2t+\ln \left|\frac(1+t)(1-t) \right|+C$ Having made the reverse substitution, we get the final result:
\[\begin(array)(l) (\int \frac((1-\sqrt(1+x+x^(2) ))^(2) )(x^(2) \sqrt(1+x +x^(2) ) dx =-2\cdot \frac(\sqrt(1+x+x^(2) ) -1)(x) +\ln \left|\frac(x+\sqrt(1 +x+x^(2) ) -1)(x-\sqrt(1+x+x^(2) ) +1) \right|+C=-2\cdot \frac(\sqrt(1+x +x^(2) ) -1)(x) +) \\ (+\ln \left|2x+2\sqrt(1+x+x^(2) ) +1\right|+C) \end (array)\]
Euler's third substitution
Under irrational understand an expression in which the independent variable %%x%% or the polynomial %%P_n(x)%% of degree %%n \in \mathbb(N)%% is included under the sign radical(from Latin radix- root), i.e. raised to a fractional power. By replacing a variable, some classes of integrands that are irrational with respect to %%x%% can be reduced to rational expressions with respect to a new variable.
The concept of a rational function of one variable can be extended to multiple arguments. If for each argument %%u, v, \dotsc, w%% when calculating the value of a function, only arithmetic operations and raising to an integer power are provided, then we speak of a rational function of these arguments, which is usually denoted %%R(u, v, \ dotsc, w)%%. The arguments of such a function can themselves be functions of the independent variable %%x%%, including radicals of the form %%\sqrt[n](x), n \in \mathbb(N)%%. For example, the rational function $$ R(u,v,w) = \frac(u + v^2)(w) $$ with %%u = x, v = \sqrt(x)%% and %%w = \sqrt(x^2 + 1)%% is a rational function of $$ R\left(x, \sqrt(x), \sqrt(x^2+1)\right) = \frac(x + \sqrt(x ^2))(\sqrt(x^2 + 1)) = f(x) $$ from %%x%% and radicals %%\sqrt(x)%% and %%\sqrt(x^2 + 1 )%%, while the function %%f(x)%% will be an irrational (algebraic) function of one independent variable %%x%%.
Let's consider integrals of the form %%\int R(x, \sqrt[n](x)) \mathrm(d)x%%. Such integrals are rationalized by replacing the variable %%t = \sqrt[n](x)%%, then %%x = t^n, \mathrm(d)x = nt^(n-1)%%.
Example 1
Find %%\displaystyle\int \frac(\mathrm(d)x)(\sqrt(x) + \sqrt(x))%%.
The integrand of the desired argument is written as a function of radicals of degree %%2%% and %%3%%. Since the least common multiple of %%2%% and %%3%% is %%6%%, this integral is an integral of type %%\int R(x, \sqrt(x)) \mathrm(d)x %% and can be rationalized by replacing %%\sqrt(x) = t%%. Then %%x = t^6, \mathrm(d)x = 6t \mathrm(d)t, \sqrt(x) = t^3, \sqrt(x) =t^2%%. Therefore, $$ \int \frac(\mathrm(d)x)(\sqrt(x) + \sqrt(x)) = \int \frac(6t^5 \mathrm(d)t)(t^3 + t^2) = 6\int\frac(t^3)(t+1)\mathrm(d)t. $$ Let's take %%t + 1 = z, \mathrm(d)t = \mathrm(d)z, z = t + 1 = \sqrt(x) + 1%% and $$ \begin(array)(ll ) \int \frac(\mathrm(d)x)(\sqrt(x) + \sqrt(x)) &= 6\int\frac((z-1)^3)(z) \mathrm(d) t = \\ &= 6\int z^2 dz -18 \int z \mathrm(d)z + 18\int \mathrm(d)z -6\int\frac(\mathrm(d)z)(z ) = \\ &= 2z^3 - 9 z^2 + 18z -6\ln|z| + C = \\ &= 2 \left(\sqrt(x) + 1\right)^3 - 9 \left(\sqrt(x) + 1\right)^2 + \\ &+~ 18 \left( \sqrt(x) + 1\right) - 6 \ln\left|\sqrt(x) + 1\right| + C \end(array) $$
Integrals of the form %%\int R(x, \sqrt[n](x)) \mathrm(d)x%% are a special case of fractional linear irrationalities, i.e. integrals of the form %%\displaystyle\int R\left(x, \sqrt[n](\dfrac(ax+b)(cd+d))\right) \mathrm(d)x%%, where %%ad - bc \neq 0%%, which can be rationalized by replacing the variable %%t = \sqrt[n](\dfrac(ax+b)(cd+d))%%, then %%x = \dfrac(dt^n - b)(a - ct^n)%%. Then $$ \mathrm(d)x = \frac(n t^(n-1)(ad - bc))(\left(a - ct^n\right)^2)\mathrm(d)t. $$
Example 2
Find %%\displaystyle\int \sqrt(\dfrac(1 -x)(1 + x))\dfrac(\mathrm(d)x)(x + 1)%%.
Let's take %%t = \sqrt(\dfrac(1 -x)(1 + x))%%, then %%x = \dfrac(1 - t^2)(1 + t^2)%%, $$ \begin(array)(l) \mathrm(d)x = -\frac(4t\mathrm(d)t)(\left(1 + t^2\right)^2), \\ 1 + x = \ frac(2)(1 + t^2), \\ \frac(1)(x + 1) = \frac(1 + t^2)(2). \end(array) $$ Therefore, $$ \begin(array)(l) \int \sqrt(\dfrac(1 -x)(1 + x))\frac(\mathrm(d)x)(x + 1) = \\ = \frac(t(1 + t^2))(2)\left(-\frac(4t \mathrm(d)t)(\left(1 + t^2\right)^2 )\right) = \\ = -2\int \frac(t^2\mathrm(d)t)(1 + t^2) = \\ = -2\int \mathrm(d)t + 2\int \frac(\mathrm(d)t)(1 + t^2) = \\ = -2t + \text(arctg)~t + C = \\ = -2\sqrt(\dfrac(1 -x)( 1 + x)) + \text(arctg)~\sqrt(\dfrac(1 -x)(1 + x)) + C. \end(array) $$
Let's consider integrals of the form %%\int R\left(x, \sqrt(ax^2 + bx + c)\right) \mathrm(d)x%%. In the simplest cases, such integrals are reduced to tabular ones if, after isolating the complete square, a change of variables is made.
Example 3
Find the integral %%\displaystyle\int \dfrac(\mathrm(d)x)(\sqrt(x^2 + 4x + 5))%%.
Considering that %%x^2 + 4x + 5 = (x+2)^2 + 1%%, we take %%t = x + 2, \mathrm(d)x = \mathrm(d)t%%, then $$ \begin(array)(ll) \int \frac(\mathrm(d)x)(\sqrt(x^2 + 4x + 5)) &= \int \frac(\mathrm(d)t) (\sqrt(t^2 + 1)) = \\ &= \ln\left|t + \sqrt(t^2 + 1)\right| + C = \\ &= \ln\left|x + 2 + \sqrt(x^2 + 4x + 5)\right| + C. \end(array) $$
In more difficult cases to find integrals of the form %%\int R\left(x, \sqrt(ax^2 + bx + c)\right) \mathrm(d)x%% are used
Ready-made answers on integrating functions are taken from the test for 1st and 2nd year students of mathematics departments. To ensure that the formulas in the problems and answers do not repeat the conditions of the tasks, we will not write out the conditions. You already know that in problems you need to either “Find the integral” or “Calculate the integral”. Therefore, if you need answers on integration, then start studying the following examples.
Integration of irrational functions
Example 18. We perform a change of variables under the integral. To simplify calculations, we select not only the root, but the entire denominator for the new variable. After such a replacement, the integral is transformed to the sum of two tabular integrals, which do not need to be simplified
After integration, we substitute a substitution for the variable.
Example 19. A lot of time and space was spent on integrating this fractional irrational function, and we don’t even know if you can figure it out from a tablet or phone. To get rid of irrationality, and here we are dealing with the cube root, we choose the root function to the third power for the new variable. Next, we find the differential and replace the previous function with the integral 
The schedule takes the most time new feature on power relations and fractions 
After the transformations, we find some of the integrals immediately, and we write the last one into two, which we transform according to the tabular integration formulas 
After all the calculations, do not forget to return to the replacement performed at the beginning 
Integrating trigonometric functions
Example 20. We need to find the integral of sine to the 7th power. According to the rules, one sine needs to be driven into a differential (we get the differential of the cosine), and the sine to the 6th power must be written through the cosine. Thus we arrive at integration from the function of the new variable t = cos (x). In this case, you will have to bring the difference to the cube, and then integrate 
As a result, we obtain a polynomial of order 7 in cosine.
Example 21. In this integral, it is necessary to write the cosine of the 4th degree in trigonometric formulas through the dependence on the cosine of the first degree. Next, we apply the tabular formula for cosine integration. 
Example 22. Under the integral we have the product of sine and cosine. According to trigonometric formulas, we write the product through the difference of sines. How this bow was obtained can be understood from the analysis of the coefficients for “x”. Next we integrate the sines 
Example 23. Here we have both a sine and a cosine function in the denominator. Moreover, trigonometric formulas will not help to simplify the dependence. To find the integral, we apply the universal trigonometric replacement t=tan(x/2)
From the record it is clear that the denominators will cancel and we will get a square trinomial in the denominator of the fraction. In it we select a complete square and a free part. After integration, we arrive at the logarithm of the difference between the prime factors of the denominator. To simplify the notation, both the numerator and denominator under the logarithm were multiplied by two. 
At the end of the calculations, instead of the variable, we substitute the tangent of half the argument.
Example 24. To integrate the function, we take the square of the cosine out of brackets, and in brackets we subtract and add one to get the cotangent. 
Next, we choose the cotangent u = ctg (x) for the new variable, its differential will give us the factor we need for simplification. After substitution we arrive at a function which, when integrated, gives the arctangent. 
Well, don’t forget to change u to cotangent.
Example 25. B last task test you need to integrate the cotangent of a double angle to the 4th degree. 
On this test integration has been decided, and not a single teacher will find fault with the answers and justification for the transformations.
If you learn to integrate like this, then tests or sections on the topic of integrals are not scary for you. Everyone else has the opportunity to learn or order solutions of integrals from us (or our competitors :))).