lecture 3 (trigonometric substitution and partial fractions)

7/31/2019 Lecture 3 (Trigonometric Substitution and Partial Fractions)

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Trigonometric Substitution Partial Fraction

Trigonometric Substitution and Integration of Rational Functions

by Partial Fractions

Mathematics 54Elementary Analysis 2

Institute of Mathematics

University of the Philippines-Diliman

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Trigonometric Substitution Partial Fraction Case1 Case 2 Case3

Trigonometric Substitution

We use trigonometric substitution for integrals involving:

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We use trigonometric substitution for integrals involving:a2u2

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We use trigonometric substitution for integrals involving:a2u2a2+u2

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b l


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We use trigonometric substitution for integrals involving:a2u2a2+u2u2

a2

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T i t i S b tit ti P ti l F ti C 1 C 2 C 3


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a2

Example:

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a2

Example:

1

49x2 dx

x

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a2

Example:

1

49x2 dx

x

2 dx

(x3)2255

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Case 1. Integrand contains the forma2u2

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go o et c Subst tut o a t a act o Case Case Case3



Let u= asin, where 2 , 2

.



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g




.

a2

u2

ua

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.

a2

u2

ua

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. Note du= acosd.

a2

u2

ua

a2

u2

=a2

a2 sin2

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. Note du= acosd.

a2

u2

ua

a2u2 =a

2a2 sin2=a2(1sin2)

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. Note du= acosd.

a2

u2

ua

a2u2 =a

2a2 sin2=a2(1sin2)= a

cos2

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. Note du= acosd.

a2

u2

ua

a2u2 =a


cos2

= a|cos|

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b


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. Note du= acosd.

a2

u2

ua

a2u2 =a


cos2

= a|cos|= acos

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C 1

2 2 L i


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Case 1.a2u2. Let u= asin.

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C 1

2 2 L t i


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Example. Find

49x2 dx

x.

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Case 1a2 u2 Let u asin


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Example. Find

49x2 dx

x.

Let x= 7sin.

49x2

x7

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Case 1a2 u2 Let u= asin


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Example. Find

49x2 dx

x.

Let x= 7sin. Then dx= 7cos d.

49x2

xdx

49x2

x7

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Case 1a2 u2 Let u= asin


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Example. Find

49x2 dx

x.


49x2

xdx

=7cos

7sin7cos d

49x2

x7

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Case 1a2u2 Let u= asin


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Case 1.a u . Let u= asin.

Example. Find

49x2 dx

x.


49x2

xdx

=7cos

7sin7cos d

49x2

x7

=

7cos2

sind

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Case 1.a u . Let u= asin.

Example. Find

49x2 dx

x.


49x2

xdx

=7cos

7sin7cos d

49x2

x7

=

7cos2

sind

=

7(1 sin2)sin

d

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Case 1.a u . Let u asin.

Example. Find

49x2 dx

x.


49x2

xdx

=7cos

7sin

7cos d

49x2

x7

=

7cos2

sind

=

7(1 sin2)sin

d

=7csc

sin d

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C

Example. Find

49x2 dx

x.


49x2

x

dx

=7cos

7sin

7cos d

49x2

x7

=

7cos2

sind

=

7(1 sin2)sin

d

=7csc

sin d

= 7 ln|csccot|+7cos+C

= 7 ln

7

x

49x2x

+

49x2+C.

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gIntegrals involving

a2u2 ,

a2+u2 or

u2 a2

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a2u2 ,

a2+u2 or

u2 a2

Case 2. Integrand contains the forma2+u2

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a2u2 ,

a2+u2 or

u2 a2


Let u= atan, where 2 , 2

.

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Integrals involvinga2u2 ,

a2+u2 or

u2 a2



. Note du= asec2d.

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a2+u2 or

u2 a2



. Note du= asec2d.

a2+u2

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a2+u2 or

u2 a2



. Note du= asec2d.

a2+u2 =a2+a2 tan2

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a2+u2 or

u2 a2



. Note du= asec2d.

a2+u2 =a2+a2 tan2

=a2(1+ tan2)

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Trigonometric Substitution2 2

2 2

2 2


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a2+u2 or

u2 a2



. Note du= asec2d.

a2+u2 =a2+a2 tan2

=a2(1+ tan2)

= a

sec2

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Trigonometric SubstitutionIntegrals involving

a2 u2

a2+u2 or

u2 a2


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a2+u2 or

u2 a2



. Note du= asec2d.

a2+u2 =a2+a2 tan2

=a2(1+ tan2)

= a

sec2

= asec a

u

a2+u2

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Case 2.a2+u2. Let u= atan.


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Example. Find

dx

(x26x+25)3.

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Example. Find

dx

(x26x+25)3.

Express x26x+25= (x3)2+42.

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Case 2. a2+u2. Let u= atan.


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Example. Find

dx

(x26x+25)3.

Express x26x+25= (x3)2+42.Let x3= 4tan. Then dx= 4sec2 d.

dx(x26x+25)3 =

dx(x3)2+42

3

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Example. Find

dx

(x26x+25)3.


4

x3

(x3)2+42

dx(x26x+25)3 =

dx(x3)2+42

3

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Example. Find

dx

(x26x+25)3.


4

x3

(x3)2+42

dx

(x26x+25)3 =

dx

(x3)2+42

3

=

4sec2

(4sec)3d

= 116

cos d

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Example. Find

dx

(x26x+25)3.


4

x3

(x3)2+42

dx

(x26x+25)3 =

dx

(x3)2+423

=

4sec2

(4sec)3d

= 116

cos d

= 116

sin+C

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Example. Find

dx

(x26x+25)3.


4

x3

(x3)2+42

dx

(x26x+25)3 =

dx

(x3)2+423

=

4sec2

(4sec)3d

= 116

cos d

= 116

sin+C

= 116

x3x26x+25

+C

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a2u2 ,

a2+u2 or

u2 a2


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a2u2 ,

a2+u2 or

u2 a2


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Case 3. Integrand contains the formu2a2

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a2u2 ,

a2+u2 or

u2 a2


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Let u= asec, where

0, 2, 32

.

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a2u2 ,

a2+u2 or

u2 a2


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Let u= asec, where

0, 2, 32

. Note du= asec tand.

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a2u2 ,

a2+u2 or

u2 a2


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Let u= asec, where

0, 2, 32


u2a2

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a2u2 ,

a2+u2 or

u2 a2


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Let u= asec, where

0, 2, 32


u2a2 =a2 sec2a2

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a2u2 ,

a2+u2 or

u2 a2


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Let u= asec, where

0, 2, 32


u2a2 =a2 sec2a2=a2(sec21)

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a2u2 ,

a2+u2 or

u2 a2


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Let u= asec, where

0, 2, 32



= a

tan2

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a2u2 ,

a2+u2 or

u2 a2


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Let u= asec, where

0, 2, 32



= a

tan2

= atan

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a2u2 ,

a2+u2 or

u2 a2


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Let u= asec, where

0, 2, 32



= a

tan2

= atan



a2u2 ,

a2+u2 or

u2 a2


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Let u= asec, where

0, 2, 32



= a

tan2

= atan a

u2a2

u

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Case 3. u2a2. Let u= asec.


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Example. Find

e4e2

ln2 x4

xlnx dx.

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l i d


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Example. Find

e4

e2ln2 x

4

xlnx dx.

Let lnx= 2sec.

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E l Fi d


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Example. Find

e4

e2ln2 x

4

xlnx dx.

Let lnx= 2sec. Then dxx= 2sec tan d.

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E l Fi d


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Example. Find

e4

e2ln2 x

4

xlnx dx.


e4e2

ln2 x4xlnx

dx

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Example Find


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Example. Find

e4

e2ln2 x

4

xlnx dx.


2

ln2 x4

lnx

e4e2

ln2 x4xlnx

dx

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Example Find


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Example. Find

e4

e2ln2 x

4

xlnx dx.


2

ln2 x4

lnx

e4e2

ln2 x4xlnx

dx=

3

0

2tan

2sec2sec tan d

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Example Find


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Example. Find

e4

e2ln2 x

4

xlnx dx.


2

ln2 x4

lnx

e4e2

ln2 x4xlnx

dx=

3

0

2tan

2sec2sec tan d

=

3

0

2tan

2sec2sec tan d

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Example Find


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Example. Find

e4

e2ln2 x

4

xlnx dx.


2

ln2 x4

lnx

e4e2

ln2 x4xlnx

dx=

3

0

2tan

2sec2sec tan d

=

3

0

2tan

2sec2sec tan d

=

3

02tan2 d

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Example Find


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Example. Find

e4

e2ln2 x

4

xlnx dx.


2

ln2 x4

lnx

e4e2

ln2 x4xlnx

dx=

3

0

2tan

2sec2sec tan d

=

3

0

2tan

2sec2sec tan d

=

3

02tan2 d

= 2

3

0(sec21) d

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Example. Find


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Example. Find

e4

e2ln2 x

4

xlnx dx.


2

ln2 x4

lnx

e4e2

ln2 x4xlnx

dx=

3

0

2tan

2sec2sec tan d

=

3

0

2tan

2sec2sec tan d

=

3

02tan2 d

= 2

3

0(sec21) d

= 2(tan)]

30

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Example. Find


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p

e4

e2ln2 x

4

xlnx dx.


2

ln2 x4

lnx

e4e2

ln2 x4xlnx

dx=

3

0

2tan

2sec2sec tan d

=

3

0

2tan

2sec2sec tan d

=

3

02tan2 d

= 2

3

0(sec21) d

= 2(tan)]

30

= 2

3 3

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Summary


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a2u2 Let u= asin

a2+u2 Let u= atan

u2a2 Let u= asec

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Trigonometric Substitution Partial Fraction Case1 Case 2 Case3 Case4

Introduction


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Introduction


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Consider 1

x2+5x+6dx.

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Introduction


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Consider 1

x2+5x+6dx.

Solvable using math 53:

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Introduction


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Consider 1

x2+5x+6dx.


1

x2+5x+6

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Introduction


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Consider 1

x2+5x+6dx.


1

x2+5x+6 =1

x2+5x+ 254 14

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Introduction


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Consider 1

x2+5x+6dx.


1

x2+5x+6 =1

x2+5x+ 254 14 =1

x+ 52

2 14

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Introduction


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Consider 1

x2+5x+6dx.


1

x2+5x+6 =1

x2+5x+ 254 14 =1

x+ 52

2 14

Note:1

x2+5x+6 =1

x+3 1

x+2

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Introduction


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Consider 1

x2+5x+6dx.


1

x2+5x+6 =1

x2+5x+ 254 14 =1

x+ 52

2 14

Note:1

x2+5x+6 =1

x+3 1

x+2The given integral will be easier to evaluate.

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We first consider rational expressions h(x)= p(x)q(x)

with deg(p(x))< deg(q(x)).

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We first consider rational expressions h(x)= p(x)q(x)

with deg(p(x))< deg(q(x)).

Goal:

To decompose such rational expression into partial fractions.

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Case 1. q(x) only has distinct linear factors.


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12/25 Trigonometric Substitution Partial Fraction Case1 Case 2 Case3 Case4


Ifq(x)= (a1x+b1)(a2x+b2) (anx+bn), then


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p(x)

q(x) =A1

a1x+b1 +A2

a2x+b2 + +An

anx+bn

whereAi

aix+bi are partial fractions.





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p(x)

q(x) =A1

a1x+b1 +A2

a2x+b2 + +An

anx+bn

whereAi


Example 1. Find

x dxx25x+6 .





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p(x)

q(x) =A1

a1x+b1 +A2

a2x+b2 + +An

anx+bn

whereAi


Example 1. Find

x dxx25x+6 .

x

x25x+6





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p(x)

q(x) =A1

a1x+b1 +A2

a2x+b2 + +An

anx+bn

whereAi


Example 1. Find

x dxx25x+6 .

x

x25x+6 =x

(x3)(x2)

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Ifq(x)= (a1x+b1)(a2x+b2) (anx+bn), then( ) A A A


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p(x)

q(x) =A1

a1x+b1 +A2

a2x+b2 + +An

anx+bn

whereAi


Example 1. Find

x dxx25x+6 .

x

x25x+6 =x

(x3)(x2) =A

x3 +B

x2 =A(x2)+B(x3)

(x3)(x2) .

Comparing the numerator of the two fractions above, we obtain

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xx25x+6 = Ax3 + Bx2


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xx25x+6 = Ax3 + Bx2

We can solve forAand Bby comparing coefficients of the equation

A( 2) B( 3) (A B) (2A 3B)


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x

=A(x

2)

+B(x

3)

=(A

+B)x

(2A

+3B).

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xx25x+6 = Ax3 + Bx2


A( 2) B( 3) (A B) (2A 3B)


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x

=A(x

2)

+B(x

3)

=(A

+B)x

(2A

+3B).

x : 1 = A+Bconstant : 0 = 2A+3B


xx25x+6 = Ax3 + Bx2


x A(x 2) B(x 3) (A B)x (2A 3B)


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x

=A(x

2)

+B(x

3)

=(A

+B)x

(2A

+3B).


Thus,A= 3 and B=2.


xx25x+6 = Ax3 + Bx2


x A(x 2) B(x 3) (A B)x (2A 3B)


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x

=A(x

2)

+B(x

3)

=(A

+B)x

(2A

+3B).


Thus,A= 3 and B=2.

Another way of solving for the values ofAand Bis to use the following substitution:


xx25x+6 = Ax3 + Bx2


x A(x 2) B(x 3) (A B)x (2A 3B)


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x

=A(x

2)

+B(x

3)

=(A

+B)x

(2A

+3B).


Thus,A= 3 and B=2.


Set x= 2:

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xx25x+6 = Ax3 + Bx2


x=A(x2)+B(x3)= (A+B)x (2A+3B)


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x=A(x 2)+B(x 3)= (A+B)x (2A+3B).


Thus,A= 3 and B=2.


Set x= 2: 2=B(1)B=2

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xx25x+6 =

Ax3 +

Bx2


x=A(x2)+B(x3)= (A+B)x (2A+3B).


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x A(x 2)+B(x 3) (A+B)x (2A+3B).


Thus,A= 3 and B=2.


Set x= 2: 2=B(1)B=2Set x= 3:

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xx25x+6 =

Ax3 +

Bx2


x=A(x2)+B(x3)= (A+B)x (2A+3B).


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Thus,A= 3 and B=2.


Set x= 2: 2=B(1)B=2Set x= 3: A= 3

Therefore,

x dx

x25x+6 =

3

x3 2

x2

dx

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xx25x+6 =

Ax3 +

Bx2


x=A(x2)+B(x3)= (A+B)x (2A+3B).


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Thus,A= 3 and B=2.


Set x= 2: 2=B(1)B=2Set x= 3: A= 3

Therefore,

x dx

x25x+6 =

3

x3 2

x2

dx= 3 ln |x3|2 ln |x2|+C.

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Case 1. Factors of the denominator are linear.


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Example 2. Find

2x34x215x+5

x22x8 dx.

Remark:

The method of partial fractions works for fractionsp(x)

q(x)with degp(x)< degq(x).

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Case 1. Factors of the denominator are linear.


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Example 2. Find

2x34x215x+5

x22x8 dx.

Since the degree of the numerator is greater than the denominator, we divide first

the numerator by the denominator. We have

2x34x215x+5x22x8

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2x3

4x2

15x

+5

x22x8 = 2x+A

x4 +B

x+2

Meanwhile,x+5

(x4)(x+2) =A

x4 +B

x+2 ,


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2x3

4x2

15x

+5

x22x8 = 2x+A

x4 +B

x+2

Meanwhile,x+5

(x4)(x+2) =A

x4 +B

x+2 ,


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x+5=A(x+2)+B(x4).

Substitute 2 for x: 3=B(6)B= 12

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2x3

4x2

15x

+5

x22x8 = 2x+A

x4 +B

x+2

Meanwhile,x+5

(x4)(x+2) =A

x4 +B

x+2 ,


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x+5=A(x+2)+B(x4).

Substitute 2 for x: 3=B(6)B= 12Substitute 4 for x:

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2x3

4x2

15x

+5

x22x8 = 2x+A

x4 +B

x+2

Meanwhile,x+5

(x4)(x+2) =A

x4 +B

x+2 ,

x+5=A(x+2)+B(x 4)


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x+5=A(x+2)+B(x4).

Substitute 2 for x: 3=B(6)B= 12Substitute 4 for x: 9=A(6)A= 32

Hence,

2x34x215x+5

x22x8 dx =

2x+ x+5(x4)(x+2)

dx

= 2x2

2+

3

2(x4) 1

2(x+2)

dx

= x2

+3

2 ln |x4|1

2 ln |x+2|+C.

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Case 2. The factors ofq(x) are all linear but some are repeated.

If (ax+b)n, n> 1 is a factor ofq(x)


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If (ax+b)n, n> 1 is a factor ofq(x) , then the partial fractions corresponding to thisfactor are A1 A2 An


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factor are A1

ax+b+A2

(ax+b)2 + +An

(ax+b)n.

Example 3. Find

(3x2) dx

2x3x2 .

3x22x3x2

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A1

ax+b+A2

(ax+b)2 + +An

(ax+b)n.

Example 3. Find

(3x2) dx2x3x2 .

3x22x3x2 =

3x2x2(2x1)

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1

ax+b+2

(ax+b)2 + +n

(ax+b)n.

Example 3. Find

(3x2) dx2x3x2 .

3x22x3x2 =

3x2x2(2x1) =

A

x+ Bx2

+ C2x1 .

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3x22x

3

x2

=A

x+B

x2

+C

2x1

Multiplying both sides byx2(2x1)

3x

2

=Ax(2x

1)

+B(2x

1)

+Cx2


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Let x= 0 B= 2Let x= 1

2 C=2

Let x= 1 A= 1

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3x22x

3

x2

=A

x+B

x2

+C

2x1

Multiplying both sides byx2(2x1)

3x

2

=Ax(2x

1)

+B(2x

1)

+Cx2


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Let x= 0 B= 2Let x= 1

2 C=2

Let x= 1 A= 1

(3x2) dx

2x3x2 =

1

x+ 2x2

22x1

dx

=ln|x|+

2x1

1 ln|2x

1|+

C

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More Examples

Example 4. Find

(x5) dx

(x+1)2

(x2).


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More Examples

Example 4. Find

(x5) dx(x+

1)2(x

2).


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x5(x+1)2(x2) =

A

x+1 +B

(x+1)2 +C

x2 .

We have x5=A(x+1)(x2)+B(x2)+C(x+1)2.

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More Examples

Example 4. Find

(x5) dx(x+

1)2(x

2).


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x5(x+1)2(x2) =

A

x+1 +B

(x+1)2 +C

x2 .

We have x5=A(x+1)(x2)+B(x2)+C(x+1)2.

Substitute 1 for x: 6=B(3)B= 2

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Case 3. Factors ofq(x) are linear and quadratic and none of the quadratic is

repeated.

Ifax2+bx+ cwhere a= 0 is a factor ofq(x) that is not repeated , then thecorresponding partial fraction to this factor is

Ax+Bax2

+bx+c

.

Example 5. Find

5 2 3 2


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5x2+3x2

x31 dx.

5x2+3x2

x31 dx=

5x2+3x2(x1)(x2+x+1) dx=

A

x1 +Bx+C

x2+x+1

dx

Consider 5x2+3x2=A(x2+x+1)+ (Bx+C)(x1).

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repeated.


Ax+Bax2

+bx

+c

.

Example 5. Find

5x2+3x 2


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5x2+3x2

x31 dx.

5x2+3x2

x31 dx=

5x2+3x2(x1)(x2+x+1) dx=

A

x1 +Bx+C

x2+x+1

dx


Substitute x= 1: 6=A(3)A= 2

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repeated.


Ax+Bax2

+bx

+c

.

Example 5. Find

5x2+3x2

d


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x31 dx.

5x2+3x2

x31 dx=

5x2+3x2(x1)(x2+x+1) dx=

A

x1 +Bx+C

x2+x+1

dx


Substitute x= 1: 6=A(3)A= 2Substitute x

=0:

2=

2(1)+C(

1)

C=

4

Substitute x=1: 0= 2(1)+ (B+4)(2)

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repeated.


Ax+Bax2

+bx

+c

.

Example 5. Find

5x2+3x2

d


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x31 dx.

5x2+3x2

x31 dx=

5x2+3x2(x1)(x2+x+1) dx=

A

x1 +Bx+C

x2+x+1

dx


Substitute x= 1: 6=A(3)A= 2Substitute x

=0:

2=

2(1)+C(

1)

C=

4

Substitute x=1: 0= 2(1)+ (B+4)(2)B= 3

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5x2+3x2x3

1

=

2x

1

+

3x+4x2

+x

+1


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5x2+3x2x3

1

= 2x

1 + 3x+4x2+x

+1

5x2+3x2

x31 dx =

2

x1 dx+

3x+4x2+x+1 dx


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= 2 ln |x1|+ 32

ln |x2+x+1|+ 52 1

3

2

tan1

x+ 12

3

2

+C

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Case 4. Some quadratic factors ofq(x) are repeated.

If (ax2+bx+ c)n, n> 1 is a factor ofq(x) , then the corresponding partial fractionsare

A1x+b1ax2

+bx

+c+ A2x+b2

(ax2

+bx

+c)2

+ + Anx+bn(ax2

+bx

+c)n


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A1x+b1ax2

+bx

+c+ A2x+b2

(ax2

+bx

+c)2

+ + Anx+bn(ax2

+bx

+c)n

Example 6. Find

x3+1


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x3+1

(x2+4)2 dx.

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A1x+b1ax2

+bx

+c+ A2x+b2

(ax2

+bx

+c)2

+ + Anx+bn(ax2

+bx

+c)n

Example 6. Find

x3+1


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x +1

(x2+4)2 dx.

x3+1

(x2+4)2 dx

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A1x+b1ax2

+bx

+c+ A2x+b2

(ax2

+bx

+c)2

+ + Anx+bn(ax2

+bx

+c)n

Example 6. Find

x3+1


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x +1

(x2+4)2 dx.

x3+1

(x2+4)2 dx=

Ax+Bx2+4 dx+

Cx+D

(x2+4)2 dx

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A1x+b1ax2

+bx

+c+ A2x+b2

(ax2

+bx

+c)2

+ + Anx+bn(ax2

+bx

+c)n

Example 6. Find

x3+1


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x +1

(x2+4)2 dx.

x3+1

(x2+4)2 dx=

Ax+Bx2+4 dx+

Cx+D

(x2+4)2 dx

x3+1= (Ax+B)(x2+4)+ (Cx+D)=Ax3+Bx2+ (4A+C)x+D.

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Continuation...

x3+1(x2+4)2 dx =

xx2+4 dx+

14x(x2+4)2 dx


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x3+1

(x2+4)2 dx=12

ln |x2+4|+ 2x2+4 +

dx(x2+4)2

Let x= 2tan


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x3+1

(x2+4)2 dx=12

ln |x2+4|+ 2x2+4 +

dx(x2+4)2

Let x= 2tan dx= 2sec2 d

x

x2+22

dx

(x2

+4)2


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2

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x3+1

(x2+4)2 dx=12

ln |x2+4|+ 2x2+4 +

dx(x2+4)2

Let x= 2tan dx= 2sec2 d

x

x2+22

dx

(x2

+4)2

=

2sec2 d

16sec4

= 18

cos2 d

= 18

1+cos2

2d


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2

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Exercises

1 Evaluate the following integrals.

1

dx

x2 +2x15

2

16e2xex

dx

34

2

x

2

4x

dx

4

x3 dx

1x25

dx

23


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25+x2

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Exercises

1 Evaluate the following integrals.

1

dx

x2 +2x15

2

16e2xex

dx

34

2

x

2

4x

dx

4

x3 dx

1x25

dx

23


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25+x2

2 Decompose the following into partial fractions

1(4x3)

x2 2x32

1

x3(x1)

3

2x

+1

x2(x2+2)

4x32x

(x2 +1)2

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lecture 3 (trigonometric substitution and partial fractions)

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