METHODS OF SOLVING SOME INDETERMINATE INTEGRALS

Volume 04 Issue 01-2024

27

American Journal Of Applied Science And Technology
(ISSN

–

2771-2745)

VOLUME

04

ISSUE

01

Pages:

27-32

SJIF

I

MPACT

FACTOR

(2021:

5.

705

)

(2022:

5.

705

)

(2023:

7.063

)

OCLC

–

1121105677

Publisher:

Oscar Publishing Services

Servi

ABSTRACT

This article shows examples of solving integrals known from the course of mathematical analysis, as well as methods
of solving given integrals using functions called Logarithmic integral function. In addition, the integral equations are
simultaneously solved by the method of integration by pieces into the differential. In turn, these types of solved
examples are very important instructions for students of mathematics, physics and engineering.

KEYWORDS

Logarithmic integral equation, indefinite integral, integration by pieces, differential, constant number, function,
hyperbolic functions.

INTRODUCTION

We know that the problem of calculating some
indefinite integrals in the course of mathematical
analysis becomes very complicated and incalculable, in
these cases we have to introduce some definitions and
solve the integrals. One such definition is the concept
of logarithmic integral functions and hyperbolic
functions. Let's first give information about the
concept of logarithmic integral function. Usually,
functions of this type are called logarithmic or integral
logarithmic functions and are denoted by the term
"li(x)". This function plays a key role in solving
examples of physics and mathematics and number

theory, because this function is very important for
approximate calculations.

𝑙𝑖(𝑥) = ∫

1

ln 𝑥

𝑑𝑥

We introduce the graph of the logarithmic integral
function and some inequalities related to it, the Maple
program was used to draw the graph, which was
certainly very useful in drawing the graph of the
function. We also get some results by comparing the
function li(x).

Research Article

METHODS OF SOLVING SOME INDETERMINATE INTEGRALS

Submission Date:

January 01, 2024,

Accepted Date:

January 05, 2024,

Published Date:

January 08, 2024

Crossref doi:

https://doi.org/10.37547/ajast/Volume04Issue01-05

Saidova Nilufar Rozimurotovna

Phd., Associate Professor Of Navoi University Of Innovations, Uzbekistan

Abdurakhmanov Gulom Erkinovich

Senior Teacher Of Navoi University Of Innovations, Uzbekistan

Journal

Website:

https://theusajournals.
com/index.php/ajast

Copyright:

Original

content from this work
may be used under the
terms of the creative
commons

attributes

4.0 licence.

Volume 04 Issue 01-2024

28

American Journal Of Applied Science And Technology
(ISSN

–

2771-2745)

VOLUME

04

ISSUE

01

Pages:

27-32

SJIF

I

MPACT

FACTOR

(2021:

5.

705

)

(2022:

5.

705

)

(2023:

7.063

)

OCLC

–

1121105677

Publisher:

Oscar Publishing Services

Servi

This function, in turn, can be compared as follows, i.e

𝑙𝑖(𝑥) = 𝑂 (

𝑥

𝑙𝑛𝑥

)

Here is an O (capital O symbol) and it looks like this

𝑙𝑖𝑥~

𝑥

ln 𝑥

∑

𝑘!

(ln 𝑥)

𝑘

∞

𝑘=0

The above comparisons help us solve calculus examples, for example the Riemann Hypothesis.

𝑙𝑖𝑥

𝑥

ln 𝑥

~1 +

1

ln 𝑥

+

2

(ln 𝑥)

2

+

6

(ln 𝑥)

3

+ ⋯

𝑙𝑖(𝑥) −

𝑥

ln 𝑥

= 𝑂 (

𝑥

(ln 𝑥)

2

)

From this we arrive at the following inequality, i.e

1 +

1

ln 𝑥

<

𝑙𝑖(𝑥) ln 𝑥

𝑥

< 1 +

1

ln 𝑥

+

3

(ln 𝑥)

2

From this inequality, ln

⁡

x≥11 for all x.

Since we used some hyperbolic functions in the calculation of the integral in the examples, we give the

following information.

Volume 04 Issue 01-2024

29

American Journal Of Applied Science And Technology
(ISSN

–

2771-2745)

VOLUME

04

ISSUE

01

Pages:

27-32

SJIF

I

MPACT

FACTOR

(2021:

5.

705

)

(2022:

5.

705

)

(2023:

7.063

)

OCLC

–

1121105677

Publisher:

Oscar Publishing Services

Servi

Hyperbolic functions

Hyperbolic functions belong to the family of elementary functions, are determined by the exponential function and
are closely related to trigonometric functions.

Below is information about these features:

Hyperbolic sine function

–

𝑠ℎ𝑥 =

𝑒

𝑥

−𝑒

−𝑥

2

;

Hyperbolic cosine function

–

𝑐ℎ𝑥 =

𝑒

𝑥

+𝑒

−𝑥

2

;

Hyperbolic tangent function

–

𝑡ℎ𝑥 =

𝑠ℎ𝑥
𝑐ℎ𝑥

=

𝑒

𝑥

−𝑒

−𝑥

𝑒

𝑥

+𝑒

−𝑥

;

Hyperbolic cotangent function

–

𝑐𝑡ℎ𝑥 =

𝑐ℎ𝑥
𝑠ℎ𝑥

=

𝑒

𝑥

+𝑒

−𝑥

𝑒

𝑥

−𝑒

−𝑥

;

A hyperbolic sequence function

–

𝑠𝑐ℎ𝑥 =

1

𝑐ℎ𝑥

=

2

𝑒

𝑥

+𝑒

−𝑥

;

Hyperbolic cosecant function

–

𝑐𝑠𝑐ℎ𝑥 =

1

𝑠ℎ𝑥

=

2

𝑒

𝑥

−𝑒

−𝑥

.

Example 1.

Calculate the given integral for all (a>0) numbers

𝑑𝑥 ∫

𝑒

𝛼𝑥

𝑥

We use the logarithmic integral function to calculate this integral.

𝑙𝑖(𝑥) = ∫

1

𝑙𝑛𝑥

𝑑𝑥.

In that case, the given integral is solved using the differential subsumption method.

∫

𝑒

𝛼𝑥

𝑥

𝑑𝑥 = ∫

1

𝛼𝑥

𝑑(𝑒

𝛼𝑥

) = ∫

1

𝑙𝑛𝑒

𝛼𝑥

𝑑(𝑒

𝛼𝑥

) = [𝑒

𝛼𝑥

= 𝑡] = ∫

1

𝑙𝑛𝑡

𝑑𝑡 =

= 𝑙𝑖(𝑡) + 𝐶 = 𝑙𝑖(𝑒

𝛼𝑥

) + 𝐶.

The use of the li(x) function plays a key role in solving the above example and is very important in calculating the
integral.

Example 2.

Volume 04 Issue 01-2024

30

American Journal Of Applied Science And Technology
(ISSN

–

2771-2745)

VOLUME

04

ISSUE

01

Pages:

27-32

SJIF

I

MPACT

FACTOR

(2021:

5.

705

)

(2022:

5.

705

)

(2023:

7.063

)

OCLC

–

1121105677

Publisher:

Oscar Publishing Services

Servi

∫ (1 −

2
𝑥

)

2

∙ 𝑒

𝑥

𝑑𝑥

When calculating this indefinite integral, we use the logarithmic integral function and the above equality

∫

𝑒

𝑥

𝑥

𝑑𝑥 =

𝑙𝑖(𝑒

𝑥

) + 𝐶

.

∫ (1 −

2
𝑥

)

2

∙ 𝑒

𝑥

𝑑𝑥 = ∫ (1 −

4
𝑥

+

4

𝑥

2

) 𝑒

𝑥

𝑑𝑥 = ∫ 𝑒

𝑥

𝑑𝑥 − 4 ∫

𝑒

𝑥

𝑥

𝑑𝑥 + 4 ∫

𝑒

𝑥

𝑥

2

𝑑𝑥 =

= 𝑒

𝑥

− 4𝑙𝑖𝑒

𝑥

− 4 ∫ 𝑒

𝑥

𝑑 (

1
𝑥

) + 𝐶 = 𝑒

𝑥

− 4𝑙𝑖𝑒

𝑥

− 4 (𝑒

𝑥

∙

1
𝑥

− ∫

𝑒

𝑥

𝑥

𝑑𝑥) + 𝐶 =

= 𝑒

𝑥

− 4𝑙𝑖𝑒

𝑥

− 4 (𝑒

𝑥

∙

1
𝑥

− 𝑙𝑖𝑒

𝑥

) + 𝐶 = 𝑒

𝑥

− 4𝑙𝑖𝑒

𝑥

− 4𝑒

𝑥

∙

1
𝑥

+ 4𝑙𝑖𝑒

𝑥

+ 𝐶 = 𝑒

𝑥

− 4𝑒

𝑥

∙

1
𝑥

+ 𝐶

Similarly, we calculate the following integral:

Example 3.

∫ (1 −

1
𝑥

) 𝑒

−𝑥

𝑑𝑥

When calculating this indefinite integral, we use the logarithmic integral function and the equality

∫

𝑒

−𝑥

𝑥

𝑑𝑥 =

𝑙𝑖(𝑒

−𝑥

) + 𝐶

shown in the first example:

∫ (1 −

1
𝑥

) ∙ 𝑒

−𝑥

𝑑𝑥 = ∫ (𝑒

−𝑥

−

𝑒

−𝑥

𝑥

) 𝑑𝑥 = ∫ 𝑒

−𝑥

𝑑𝑥 − ∫

𝑒

−𝑥

𝑥

𝑑𝑥 = 𝑒

−𝑥

− 𝑙𝑖𝑒

−𝑥

+ 𝐶

In addition, the logarithmic integral function is also important for solving the following examples, it can be seen from
the examples solved above that

∫

𝑒

𝛼𝑥

𝑎𝑥

2

+ 𝑏𝑥 + 𝑐

𝑑𝑥 (𝑏

2

− 4𝑎𝑐 ≥ 0) , ∫

𝑒

𝛼𝑥

𝑎𝑥 + 𝑏

(𝑎 ≠ 0)

integrals of this form can be calculated using the logarithmic integral function.

Example 4.

Calculate the given integral:

∫

𝑒

2𝑥

𝑥

2

− 3𝑥 + 2

𝑑𝑥

Volume 04 Issue 01-2024

31

American Journal Of Applied Science And Technology
(ISSN

–

2771-2745)

VOLUME

04

ISSUE

01

Pages:

27-32

SJIF

I

MPACT

FACTOR

(2021:

5.

705

)

(2022:

5.

705

)

(2023:

7.063

)

OCLC

–

1121105677

Publisher:

Oscar Publishing Services

Servi

When calculating this indefinite integral, we also rely on the equality shown in Example 1:

∫

𝑒

2𝑥

𝑥

2

− 3𝑥 + 2

𝑑𝑥 = ∫ 𝑒

2𝑥

(

1

𝑥 − 2

−

1

𝑥 − 1

) 𝑑𝑥 = 𝑒

4

∫

𝑒

2(𝑥−2)

𝑥 − 2

𝑑(𝑥 − 2) −

−𝑒

2

∫

𝑒

2(𝑥−1)

𝑥 − 1

𝑑(𝑥 − 1) = 𝑒

4

∙ 𝑙𝑖𝑒

2(𝑥−2)

− 𝑒

2

∙ 𝑙𝑖𝑒

2(𝑥−1)

+ 𝐶

Similarly, we calculate the following integrals:

Example 5.

Calculate the given integral:

∫

𝑥𝑒

𝑥

(𝑥 + 1)

2

𝑑𝑥

Using the equation in Example 1, we solve:

∫

𝑥𝑒

𝑥

(𝑥 + 1)

2

𝑑𝑥 = ∫

(𝑥 + 1)𝑒

𝑥

(𝑥 + 1)

2

𝑑𝑥 − ∫

𝑒

𝑥

(𝑥 + 1)

2

𝑑𝑥 =

= ∫

𝑒

𝑥+1

𝑥 + 1

𝑑(𝑥 + 1) + ∫ 𝑒

𝑥

𝑑 (

1

𝑥 + 1

) = 𝑙𝑖(𝑒

𝑥+1

) + 𝑒

𝑥

∙

1

𝑥 + 1

−

1
𝑒

∫

𝑒

𝑥+1

𝑥 + 1

𝑑(𝑥 + 1 ) + 𝐶

= 𝑙𝑖(𝑒

𝑥+1

) +

𝑒

𝑥

𝑥 + 1

−

1
𝑒

∙ 𝑙𝑖(𝑒

𝑥+1

) + 𝐶 =

𝑒

𝑥

𝑥 + 1

+ (1 −

1
𝑒

) 𝑙𝑖(𝑒

𝑥+1

) + 𝐶

Thus, we use the logarithmic integral function to solve the following final integral.

∫ 𝑒

2𝑥

∙

𝑥

4

(𝑥 − 2)

2

𝑑𝑥 = ∫ 𝑒

2𝑥

(𝑥

2

+ 4𝑥 + 12 +

32

𝑥 − 2

+

16

(𝑥 − 2)

2

) 𝑑𝑥 =

= ∫ 𝑒

2𝑥

(𝑥

2

+ 4𝑥 + 12)𝑑𝑥 + 32𝑒

4

∫

𝑒

2(𝑥−2)

𝑥 − 2

𝑑𝑥 − 16 ∫ 𝑒

2𝑥

𝑑 (

1

𝑥 − 2

) =

=

𝑒

2𝑥

2

(𝑥

2

+ 3𝑥 +

21

2

) + 32𝑒

4

𝑙𝑖𝑒

2(𝑥−2)

− 16 (𝑒

2𝑥

∙

1

𝑥 − 2

− 2𝑒

4

∫

𝑒

2(𝑥−2)

𝑥 − 2

𝑑(𝑥 − 2)) + 𝐶 =

=

𝑒

2𝑥

2

(𝑥

2

+ 3𝑥 +

21

2

) + 32𝑒

4

𝑙𝑖(𝑒

2(𝑥−2)

) − 16𝑒

2𝑥

∙

1

𝑥 − 2

+ 32𝑒

4

𝑙𝑖(𝑒

2(𝑥−2)

) + 𝐶 =

=

𝑒

2𝑥

2

(𝑥

2

+ 3𝑥 +

21

2

− 32) + 64𝑒

4

𝑙𝑖(𝑒

2(𝑥−2)

) + 𝐶

CONCLUSION

So, in the calculation of integrals, we come across
certain types of integrals, and in the calculation of

these types of integrals, we used notations that bring
us convenience. In addition, we use the li(x) function to
calculate

complex

integrals

of

the

form

Volume 04 Issue 01-2024

32

American Journal Of Applied Science And Technology
(ISSN

–

2771-2745)

VOLUME

04

ISSUE

01

Pages:

27-32

SJIF

I

MPACT

FACTOR

(2021:

5.

705

)

(2022:

5.

705

)

(2023:

7.063

)

OCLC

–

1121105677

Publisher:

Oscar Publishing Services

Servi

∫

𝑒

𝛼𝑥

𝑎𝑥

2

+𝑏𝑥+𝑐

𝑑𝑥 (𝑏

2

− 4𝑎𝑐 ≥ 0)

.

In

addition,

the

logarithmic integral function makes it much easier for
students to calculate this type of integrals.

REFERENCES

1.

B.P. Demidovich. Collection of problems and
exercises in mathematical analysis. Publishers in
1970. Pages: 497

2.

Alimov Sh.O., Ashurov R.R. Mathematical analysis.
Part 1. "Rainbow" publishing house/ Tashkent 2012.

3.

Popov I.Yu. Problems of increased difficulty in the
course of higher mathematics. St. Petersburg,
2008.

4.

Ambartsumyan

B.A.,

Andryushchenko

E.A.,

Bkhensky K.V. and others. Student Mathematical
Olympiads. Part 1. Ryazan. 2014