Integration by Partial Fraction
26
Intregation of Rasional Functions by Partial Fractions
-
Upload
syarifah-nabila-aulia-ranti -
Category
Documents
-
view
15 -
download
0
description
MODUL CALCULUS 2UNIVERSITAS ISLAM INDONESIA
Transcript of Integration by Partial Fraction
Intregation of Rasional Functions by Partial Fractions
Intregation of Rasional Functions by Partial Fractions
In general, then, we are concerned with the integration of expressions of the form:
dxxQ
xP
)(
)(
)(
)()(
xQ
xPxf
• There are 2 types of rational function:1. Proper Rational function: if the
degree of the numerator P(x) is less than the degree of denominator Q(x). ex. :
2. Improper Rational Function: if the degree of the numerator P(x) greater or equal than degree of denominator Q(x) ex.:
Rational Function
84
22)(
2
xx
xxf
xx
xxf
3
3 1)(
xx
xxxxg
5
12)(
3
35
2)1(
53)(
x
xxf
To do this technique, it is necessary to write P(x)/Q(x) as the sum of partial fraction.
The denominator of the partial fractions are obtained by factoring Q(x) into a product of linear and quadratic factors.
The method of determining the partial fraction depends on the nature of these factors.
Integration by Partial Fraction
dxcx
C
bx
B
ax
Adx
cxbxax
xP
))()((
)(
dxbx
C
ax
B
ax
Adx
bxax
xP
22 )()()(
)(
dxcbxax
CBx
ax
Adx
cbxaxax
xP
22 ))((
)(
dxrqxpx
EDx
cbxax
CBxdx
rqxpxcbxax
xP
2222 ))((
)(
Take these steps to evaluate the integral Step 1 : Factor the denominator as much as possible to get the form of the partial fraction
decompositionStep 2 : Multiply both sides by the denominator and
then set the numerators are equal for every x to get the constantsStep 3 : Plug the constant to the above equation,
and evaluate the integral
Examples
Example 1.1. Faktor linier berbeda
2. Faktor linier berulang
dx
xx
x
6
13.1
2
20
)12)(3(
19 2. dx
xx
x
Exercises
Find the integral!
dxxxx
dxxx
x
dxxx
x
2
1 23
2
2
0
168
1 .3
)4()2( .2
)12)(3(
19 .1
2
0
)12)(3(
19 .1 dx
xx
x
Answer : dxx
B
x
Adx
xx
x
123
)12)(3(
19 2
0
2
0
)12)(3(
)3()12(
)12)(3(
19
123)12)(3(
19
xx
xBxA
xx
x
x
B
x
A
xx
x
7
26 726
)16(1)3(9 3for
1 2
13
2
13
32
101
2
19
2
1for
)3()12(19Then
AA
Ax
BB
Bx
xBxAx
2
0
2
0
2
0
2
0
2
0
2
0
22ln2
13ln
7
26
12
1
3
1
7
26
12
1
3
7/26
123
)12)(3(
19
xx
dxx
dxx
dxxx
dxx
B
x
A
xx
x
Then ->
89,1350
663
)12)(3(
192
0
dxxx
x
)69,0(2
1)1,1(
7
26)69,0(
2
1)61,1(
7
26
20.2ln2
130ln
7
2622.2ln
2
132ln
7
26
350
66353
143
50
299
500
69
53
143
200
69
50
299
dx
xx
x
)4()2( .2
2
Answer : dxx
C
x
B
x
Adx
xx
x
4)2(2)4()2( 22
)4()2(
)2()4()4)(2(
)4()2(
4)2(2)4()2(
2
2
2
22
xx
xCxBxxA
xx
x
x
C
x
B
x
A
xx
x
9
1 8
9
8
9
4
3
480
)2(4)4)(2(0 0For
9
1 364
)24(004 4For
3
1 6 2
0)42(02 2For
)2()4()4)(2(Then
2
2
2
AA
A
CBAx
CC
Cx
BB
Bx
xCxBxxAx
Cxxx
dxxxx
dxx
C
x
B
x
Adx
xx
x
4ln9
1)2(
3
12ln
9
1
4
9/1
)2(
3/1
2
9/1
4)2(2)4()2(
1
2
22
Cxxxdxxx
x
4ln
9
1)2(
3
12ln
9
1
)4()2(
19 12
dxxxx
2
1 23
168
1 .3
Answer : dxx
C
x
B
x
Adx
xx
)4()4(
)4(
1 2
1 2
2
1 2
2
2
2
22
)4(
)4()4(
)4(
1
)4()4()4(
1
xx
CxxBxxA
xx
x
C
x
B
x
A
xx
16
1 3
16
3
4
13
16
91
391 1For
4
1 41
4001 4For
16
1 16 1
0016 1 0For
)4()4(1then 2
BA
B
CBAx
CC
Cx
AA
Ax
CxxBxxA
2
1
2
1 2
2
1 2
2
1 2
)4(4
14ln
16
1ln
16
1
)4(4
1
)4(16
1
16
1
)4()4(
)4(
1
xxx
dxxxx
dxx
C
x
B
x
Adx
xx
then ->
480
5312
1
160
11
8
1
)12(
1)1,1(
16
1)0(
16
1
)8(
1)69,0(
16
1)69,0(
16
1
)41(4
141ln
16
11ln
16
1
)42(4
142ln
16
12ln
16
1
11,0480
53
168
12
1 23
dxxxx
