Integrating Rational Functions by the Method of Partial Fraction.
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Integrating Rational Functions by the Method of Partial Fraction
Examples I
When the power of the polynomial of the numerator is less than that
of the denominator
Example 1
dxxxx
x
)12)(1(
4822
)12)(1(
48)(
22
xxx
xxr
Let
)12)(1(
)()2()2()(
)12)(1(
)12)(()1()1)(1(
)12)(1(
)1)(()1()1)(1(
1)1(1)(
)1)(1(
)12)(1(
22
23
22
222
22
222
22
22
22
xxx
dbaxdcaxdcbaxca
xxx
xxdcxxbxxa
xxx
xdcxxbxxa
x
dcx
x
b
x
axr
xx
xxx
The denominator
8
42
02
0
,
)()2()2()(
4823
dba
dca
dcba
ca
getWe
dbaxdcaxdcbaxca
numeratornewtheandxnumeratororiginaltheComparing
246,4)4(
482022)6(
:),2(
682
:),4(
242
42
:),3(
)1(
)4(8
)3(42
)2(02
)1(0
bc
aaaaa
getweinthatngSubstituti
abba
getweinthatngSubstituti
dd
daa
getweinthatngSubstituti
acFrom
dba
dca
dcba
ca
cxxx
x
dxx
x
xx
dxxr
Thus
x
x
xx
x
dcx
x
b
x
axr
d
c
b
a
getweequationslinearofsystemthisSolving
arctan2)1ln(21
)1(21ln4
]1
24
)1(
2
1
4[
)(
,
1
24
)1(
2
1
4
1)1(1)(
2
4
2
4
,,
21
22
22
22
Example 2
xx
dx2
)1(
11)(
,
11
,,
10
,,
)1(
)(
)1(
)1(
)1()1(
1
)1(
11)(
2
xxxr
Thus
banda
getweequationslinearofsystemthisSolving
aandba
getwenumeratorstwotheComparing
xx
axba
xx
bxxa
x
b
x
a
xx
Let
xxxxxr
cx
x
cxx
dxxx
dxxr
Thus
1ln
1lnln
])1(
11[
)(
,
Example 3
dxx
xx
22
2
)1(
12
01,2
1,2,1,0
,
)()(
)()1)((
)1()1(
)1(
12)(
)1(
12
23
2
222
22
2
22
2
bdc
dbcaba
SolvingandnumeratorsComparing
dbxcabxax
dcxxbax
numeratornewThe
x
dcx
x
bax
x
xxxr
dxx
xx
cx
x
cx
x
dxx
x
x
dxxr
1
1arctan
1
)1(arctan
])1(
2
)1(
1[
)(
2
12
222
Examples II
When the power of the polynomial of the numerator is equal or greater than
that of the denominator
Example 1
dxx
x
1
22
cxxx
dxx
xdxx
x
Thus
xx
x
x
x
x
methodAlternativ
xx
x
xxr
getweDividing
dxx
x
1ln32
]1
31[
1
2
,
1
31
1
3)1(
1
2
:
1
31
1
2)(
,,1
2
2
2
22
2
2
Notice
We can use the same method used in this example as an alternative way to write the given rational function as a sum of simpler rational functions (partial fractions).
Going back to examples. Notice the following:
)1(
11
)1()1(
1
)1(
)1(
)1(
1
1
2
2
xx
xx
x
xx
x
xx
xx
xx
xx
Example
222
22
2
22
2
)1(
2
1
1
)1(
2)1(
)1(
12
3
x
x
x
x
xx
x
xx
Example
Question( Use two methods)
12x
dx
Example 2
12
3
x
dxx
1ln2
1
2)(
1
1
)(
1)(
1
22
2
2
3
2
3
2
3
xx
dxxr
x
xx
x
xxx
x
xxr
x
dxx
Examples III
Sometimes it is easier to find the constants not by solving a system of linear equations but rather by substituting a different appropriate value for x, in each of these equationsز
Example 1
dxxxxxxx
x
)5)(4)(3)(2)(1(
12015
54321
)5)(4)(3)(2)(1(
12015)(
)5)(4)(3)(2)(1(
12015
543210
x
a
x
a
x
a
x
a
x
a
x
a
xxxxxx
xxr
dxxxxxxx
x
)4)(3)(2)(1(
)5)(3)(2)(1(
)5)(4)(2)(1(
)5)(4)(3)(1(
)5)(4)(3)(2(
)5)(4)(3)(2)(1(
5
4
3
2
1
0
xxxxxa
xxxxxa
xxxxxa
xxxxxa
xxxxxa
xxxxxa
numeratornewThe
1)5)(4)(3)(2(
120
120)5)(4)(3)(2)(1(
,
120120)0(15
)40)(30)(20)(10)(0(
)50)(30)(20)(10)(0(
)50)(40)(20)(10)(0(
)50)(40)(30)(10)(0(
)50)(40)(30)(20)(0(
)50)(40)(3)(20)(10(
0,
0
0
5
4
3
2
1
0
a
a
Thus
numeratororiginalThe
a
a
a
a
a
xa
numeratornewThe
xSubstitute
8
35
)4)(3)(2(
105
105)4)(3)(2)(1)(1(
,
105120)1(15
)51)(41)(31)(21)(1(
1
1
1
1
a
a
Thus
numeratororiginalThe
a
numeratornewThe
xSubstitute
4
25
)2)(1)(1)(2)(3(
75
120)3(15)2)(1)(1)(2)(3(
,,3
2
15
)3)(2)(1)(1)(2(
90
120)2(15)3)(2)(1)(1)(2(
,,2
3
3
2
2
a
a
getwexSubstitute
a
a
getwexSubstitute
8
3
)1)(2)(3)(4)(5(
45
120)5(15)1)(2)(3)(4)(5(
,,5
2
5
)1)(1)(2)(3)(4(
60
120)4(15)1)(1)(2)(3)(4(
,,4
5
5
4
4
a
a
getwexSubstitute
a
a
getwexSubstitute
cxx
xxxx
dxxxxxxx
dxxr
Therfore
5ln8
34ln
2
5
3ln2
252ln
2
151ln
8
35ln
]583
425
3225
2215
1835
1[
)(
,
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