mat 1235 calculus ii section 7.4 partial fractions
TRANSCRIPT
MAT 1235Calculus II
Section 7.4
Partial Fractions
http://myhome.spu.edu/lauw
HW
Please do your HW ASAP. Please actually do your HW.
Partial Fractions
Review PF from pre-calculus Use PF to simplify integrands Break up a complicated rational function
into smaller ones Each of the smaller rational function is
easier to integrate
Preview
Use Partial Fractions to decompose a rational function into a sum of simpler rational functions
dxxQ
xPdx
xQ
xPdx
xQ
xPdx
xQ
xP
)(
)(
)(
)(
)(
)(
)(
)(
2
2
2
2
1
1
Assumption
We assume: deg(P(x))<deg(Q(x))
If this is not the case, we can use long division to rewrite the rational function as
)(
)()(
)(
)(
xQ
xRxS
xQ
xP
1
11
12
3
xxx
x
x
( )
( )
P xdx
Q x
2
1
xdx
x 2
3
3xdx
x x
Assumption
We assume: deg(P(x))<deg(Q(x))
If this is not the case, we can use long division to rewrite the rational function as
)(
)()(
)(
)(
xQ
xRxS
xQ
xP
1
11
12
3
xxx
x
x
( )
( )
P xdx
Q x
Quotient
Remainder
Example 1 2
1
xdx
x
)(
)()(
)(
)(
xQ
xRxS
xQ
xP
Remark
In stead of using the substitution, we can use the following formula
Caxdxax
ln1
Example 2
1
xdx
x
Assumption
We assume: deg(P(x))<deg(Q(x)) Depends on the form of Q(x), we have 3
different cases.
( )
( )
P xdx
Q x
Case I
Q(x) is a product of distinct linear factors
( 1)( 2)
5xd
x xx
Case I
Q(x) is a product of distinct linear factors
kk
k
kk
bxa
A
bxa
A
bxa
A
xQ
xR
bxabxabxaxQ
22
2
11
1
2211
)(
)(
)())(()(
( 1)( 2)
5xd
x xx
( 1)( 2) 2
5
1
x A B
x x x x
Example 3
dxxx
x
)2)(1(
5
: Compare coefficeints
: Plug in numbers
Method I
Method II
( 1)( 2) 2
5
1
x A B
x x x x
Expectation
Make sure you write down the final partial fractions (on the right hand side) before you proceed to evaluate the integral.
To save time…
We will work on the partial fractions only We are not going to actually complete
the integration
Case II
If (axi+bi) is repeated r times
2
1
(( 1) 2)xdx
x
Case II
If (axi+bi) is repeated r times, we use
1 2
2r
ri i i i i i
A A A
a x b a x b a x b
( ) ( )ri iQ x a x b
2 2
1
( 2) 2( 1) 1 ( 1)
C
x x
A B
x x x
2
1
(( 1) 2)xdx
x
Example 4
dx
xx )2()1(
12
Example 4
dxxx )2()1(
12
2 2
2
1
( 1) ( 2) 1 ( 1) 2
1 ( 1)( 2) ( 2) ( 1)
A B C
x x x x x
A x x B x C x
Example 4
dxxx )2()1(
12
2 2
2
2 2
2
1
( 1) ( 2) 1 ( 1) 2
1 ( 1)( 2) ( 2) ( 1)
( 1) ( 2) ( 1) ( 2)
1 ( 1)( 2) ( 2) ( 1)
A B C
x x x x x
A x x B x C x
x x x x
A x x B x C x
Example 4
dxxx )2()1(
12
CBA
xCxBxxA
,,for Solve
)1()2()2)(1(1 2
2 2
1
( 1) ( 2) 1 ( 1) 2
A B C
x x x x x
Case III
If Q(x) has a irreducible factor ax2+bx+c,
2( 1
1
)xdxx
Case III
If Q(x) has a irreducible factor ax2+bx+c,
we use
2
Ax B
ax bx c
2( 1
1
)xdxx 2 2
1
( 1) 1
Ax
x
C
x
B
x x
Example 5
2
1
( 1)dx
x x 2 2
1
( 1) 1
Ax
x
C
x
B
x x