Section 41: Indefinite Integrals · Properties of the Indefinite Integral: to as many functions as we need. difference of the individual integrals.This rule can be extended the integral

Definition

Section 4.1: Indefinite Integrals

The process of finding the indefinite integral is called integrationor integrating f(x). If we need to be specific about the integration variable we will say that we are integrating f(x) with respect to x.

Properties of the Indefinite Integral:

need. weas functionsmany as to

extended becan rule Thisintegrals. individual theof difference

or sum theis functions of differenceor sum a of integral the

s,other wordIn 2 .)()()()( ∫∫∫ ±=±− dxxgdxxfdxxgxf

integrals. indefinite ofout constants tivemultiplicafactor can

weSo, number.any iswhere1 ,)()(. kdxxfkdxxfk ∫∫ =−

.)()()()( ∫∫∫ ×≠× dxxgdxxfdxxgxf

.)(

)(

)()(

∫∫

∫ ≠dxxg

dxxfdx

xgxf

.1,1

1

−≠++

=+

∫ ncnxdxxn

n

.constants areand, kcckxkdx +=∫

.cossin cxxdx +−=∫ .sincos cxxdx +=∫

.tansec2 cxxdx +=∫ .sectansec cxxdxx +=∫

The first integral that we will look at is the integral of a power of x.

The general rule when integrating a power of x we add one onto the exponent and then divide by the new exponent. It is clear that we will need to avoid n = -1 in this formula. If we allow in this formula we will end up with division by zero.

Next is one of the easier integrals but always seems to cause problems for students.

Let us now take a look at the trigonometric functions:

Example:

.cedxe xx +=∫ .||ln1 1 cxdxxdxx

+== ∫∫ −

.4105)( 63∫ +− − dttta

.)( 88∫ −+ dxxxb

.42454

510

45

4105

5454

63

ctttcttt

dtdttdtt

+++=++−

−

=+−=

−−

−∫ ∫∫

.7979

797988 cxxcxxdxxdxx +−=+

−+=+=

−−−∫∫

.1 cydy +== ∫.)( ∫dyc

Now, let us take care of exponential and logarithm functions.

.6773)( 5

4 3∫ ++ dxxx

xd ∫∫∫ ++= dxx

dxx

dxx 167173 5

4 3

( )∫ ∫ ∫−− ++= dxxdxxdxx 21

541

3

6773

cxxx+

+−+

+−+

+=

+−+−+

1216

715

71

433

121

15143

cxxx++−=

−

37

47

712 2

144

7

( )( ) .4)( 23∫ −+ dwwwwe

.1524)( 3

2610

∫+− dx

xxxxf

( ) .44 3233∫ −+−= dwwwwww

∫∫∫ −+−= dxxdxxdxx 137 1524

.44 37

31

3∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛−+−= dwwwww

.

310

34442

4310

34

42

cwwww+−+−=

.1033

42

310

344

2 cwwww +−+−=

cxxx++−= ||ln15

42

84

48

cxxx++−= ||ln15

22

48

Integration by substitution:

∫ ∫∫∫ dyewdwdtt

dxx ycos13

4

( ) ( )∫ ∫ −−−⎟⎠

⎞⎜⎝

⎛ − dyeydwwww

yy2418lncos11

∫∫ +

++ dt

tttdxxx 34

34 32

)2(125618

After the last section we now know how to do the following integrals

However, we can’t do the following integrals:

Let us start with the first one

∫ + dxxx 4 32 5618

In this case let us notice

( ) ( ) ∫∫∫ =+=+ duudxxxdxxx 41

241

34 32 18565618

56 3 += xu

and we compute the differential dxxdxudu 218' ==

Now, we go back to our integral and notice that we can eliminate every x that exists in the integral and write the integral completely in terms of u using both the definition of u and its differential

Evaluating the integral gives,

( ) cxcuduudxxx ++

=+==+ ∫∫ 5564

545618

45

345

41

4 32

Example:

Solution:

( )∫ −⎟⎠

⎞⎜⎝

⎛ − dwwww

a lncos11)(

( )∫ − dxxxc 432 103)(

( )∫ −− dyeyb yy24183)(

∫−

dxx

xd241

)(

(a) In this case let us take wwu ln−=

and we compute the differential dw

wdwudu ⎟

⎠

⎞⎜⎝

⎛ −==11'

Now, we go back to our integral and notice that we can eliminate every w that exists in the integral and write the integral completely in terms of u using both the definition of u and its differential

( ) ( )

( ) ( ) cwwcuduu

dww

wwdwwww

+−=+==

⎟⎠

⎞⎜⎝

⎛ −−=−⎟⎠

⎞⎜⎝

⎛ −

∫

∫∫

lnsin)sin(cos

11lncoslncos11

( )∫ −− dyeyb yy24183)( In this case let us take yyu −= 24and we compute the differential ( )dyydyudu 18' −==

Thus we get

( ) ( ) ceceduedyyedyey yyuuyyyy +=+==−=− −−− ∫∫∫222 444 333183183

In this case let us take 3103 xu −=

So, dxxdxudu 230' −==

Thus we get( ) ( )∫∫ −−

−=− dxxxdxxx )30(103301103 243432

( )∫ − dxxxc 432 103)(

( ) cxcuduu +−−

=+−

=−

= ∫ 150103

1501

301 53

54

∫−

dxx

xd241

)(

In this case let us take 241 xu −=

So,( )dxxdxudu 8' −==

Thus,

( )∫∫∫ −−−=−=−

−−dxxxxdxxdx

xx 8)41(

81)41(

4121

221

2

2

cuduu +−=−= ∫−

218

181 2

1

21

cx +−−= 21

2 )41(41

Integration by parts:

Example 1:

Solution:∫ dxxex Find

∫ dxex x2 FindExample 2:

Solution:

.,., Then Let xx evdxdudxedvxu ====

c. +−=−= ∫∫ xxxxx exedxexedxxe

.,2., Then Let 2 xx evxdxdudxedvxu ====

∫∫ −= dxxeexdxex xxx 222

( )c22 +−−= xxx exeex

C222 ++−= xxx exeex

Inserting this integral in the first one yields

Bringing the last term to the left hand side and dividing by 2 gives

Example 3:

Solution:∫ xdxex cos Find

.sin,.cos, Then Let xvdxeduxdxdveu xx ====

.sinsincos Thus, ∫∫ −= xdxexexdxe xxx

So,Then Let

. integral theagain partsby integrate Now

.cos,.sin,

sin

xvdxeduxdxdveu

xdxexx

x

−====

∫

.coscos)cos(cossin ∫∫∫ +=−−= xdxexedxxexexdxe xxxxx

.coscossincos ∫∫ −−= xdxexexexdxe xxxx

xexexdxe xxx cossincos2 −=∫( ) .cossin21cos cxexexdxe xxx +−=∴∫

Example 4:

Solution:.

5,1.,ln

54 Then Let

xvdxx

dudxxdvxu ====

.15

ln5

ln Thus,55

4 ∫∫ −= dxx

xxxxdxx

.ln Find 4∫ xdxx

∫−= dxxxx 45

51ln

5

c55

1ln5

55

+−=xxx

c.25

ln5

55

+−=xxx

Example 5:

Solution:.,1..1,ln Then Let xvdx

xdudxdvxu ====

.1lnln Thus, ∫∫ −= dxxxxxxdx

.ln Find ∫ xdx

∫−= dxxx .1ln

cln +−= xxx

Integration by partial fractions:

Main Rules:

Example 1:

First we write

Thus

Integrating, we get

Distinct roots:

∫ −− )2)(1( Integrate

xxxdx

Solution:

21)2)(1( −+

−=

−− xB

xA

xxx

getting ,)2)(1(by equation hismultiply t Now x-x-

)1()2( −+−= xBxAx;1 so ,)21(1get we,1 substitute weIf -AAx =−==

.2 so ,)12(2get we,2 letting Now =−== BBx

22

11

)2)(1( −+

−

−=

−− xxxxx

.|2|ln2|1|ln)2(

2)1()2)(1(

cxxxdx

xdx

xxxdx

+−+−−=−

+−

−=−− ∫∫∫

Integrating, we obtain

Therefore

Example 2: ∫ −−

−

)3)(1()3( Integrate 2

2

xxdxx

Solution:

)3)(1)(1( getting ,by equation hismultiply t Now −+− xxx

;4/1 so ,)4)(2(3-1get we,1 Substitute -AAx =−−=−=

and writeFirst we ),1)(1()1( 2 +−=− xxx

)3()1()1()3)(1)(1()3(

)3)(1()3( 2

2

2

−+

−+

+=

−+−

−=

−−

−

xD

xB

xA

xxxx

xxx

)1)(1()3)(1()3)(1()3( 2 −++−++−−=− xxDxxBxxAx

;2/1 so ,)2)(2(3-1get we,1 Substitute =−== BBx;4/38/6 so ,)2)(4(3-9get we,3 Substitute ==== CCx

)3(1

43

)1(1

21

)1(1

41

)3)(1()3(

2

2

−+

−+

+−=

−−

−

xxxxxx

.|3|ln43|1|ln

21|1|ln

41

)3)(1()3(

2

2

cxxxxxdxx

+−+−++−=−−

−∫

which we can integrate term by term

Thus

Therefore

Multiple roots:

Example 3:

First we write

∫ −

+

)1()1( Integrate 2

2

xxdxx

Solution:1)1(

)1(22

2

−++=

−

+

xD

xB

xA

xxx

1211

)1()1(

22

2

−+−−=

−

+

xxxxxx

22 )1()1(1 DxxBxAxx +−+−=+

.1 so ,)1(1get we,0 ngSubstituti −=−== BBx

; of eother valuany take we find To .2get we,1 ngSubstituti xADx ==

.1soand,22-

so ,2)2)(1()2)(1(2get we,1 Substitute

−==

+−−+−−=−=

AA

Ax

.|1|ln21||ln)1()1(

2

2

cxx

xxxdxx

+−++−=−

+∫

Example 4:Solution:

First we write

Multiply by we get

Thus

Integrating, we obtain

83,So.

89

885121

85

233

,85

233331get we0 Substitute

==−+

=−+=

++−=++−==

AA

ADBA x

3)1(1)1)(3()1(

22

2

++

−+

−=

−+

+

xD

xB

xA

xxx

22 )1()3()1)(3(1 −+++−+=+ xDxBxxAx

.85,)4(19get we,3 Substitute 2 =−=+−= DsoD x

.21),4(2get we,1 Substitute === BsoB x

31

85

)1(1

21

11

83

)1)(3()1(

22

2

++

−+

−=

−+

+

xxxxxx

.|3|ln85

)1(21|1|ln

83

)1)(3()1(

2

2

cxx

xxxdxx

+++−

−−=−+

+∫

Solution:

Example 5:

No roots:Main Rule:

Thus

cxxdx

xdx

+=+

=+

−∫∫ )3(tan

31

391

222

cax

aaxdx

+=+

−∫ )(tan1 122

∫ +9 Evaluate 2x

dx

Solution:

Example 6:

1)2(45)44(54 222 +−=−++−=+− xxxxx

∫ +− 54 Evaluate 2 xx

dx

Here we can’t find real factors, because the roots are complex. But we can complete the square:

.)2(tan1)2(54

122 cx

xdx

xxdx

+−=+−

=+−

−∫∫

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