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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 17 "One more e
xample." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
79 "Integrate the following function with the methods from the previou
s worksheets." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "restart: f
 := (-4*x^9+x^6+16*x^5+22*x^3+4*x^2+6*x+1)/(x^8+3*x^6+3*x^4+x^2):" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 89 "First we have to normalize f, then
 take the square-free factorization of the denominator:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "f := normal(f); v := denom(f); v :=
 sqrfree(v,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG,$*&,0*$)%\"xG
\"\"*\"\"\"\"\"%*$)F*\"\"'F,!\"\"*$)F*\"\"&F,!#;*$)F*\"\"$F,!#A*$)F*\"
\"#F,!\"%F*!\"'F1\"\"\"F,*&)F*\"\"#F,,*F.F?*$)F*F-F,F8F:F8F?F?\"\"\"!
\"\"F1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"vG*&)%\"xG\"\"#\"\"\",**
$)F'\"\"'F)\"\"\"*$)F'\"\"%F)\"\"$*$F&F)F2F.F.F." }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"vG7$\"\"\"7$7$%\"xG\"\"#7$,&*$)F)F*\"\"\"F&F&F&\"\"
$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Highest pole order = 3.  Thi
s pole order is located at:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
11 "d := x^2+1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG,&*$)%\"xG\"
\"#\"\"\"\"\"\"F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "Now if A1
 is some known function, then integrating f can be reduced to: integra
ting f1 where f1 = f - A1'" }}{PARA 0 "" 0 "" {TEXT -1 81 "Why? Becaus
e the antiderivative of f  is simply A1 plus the antiderivative of f1.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 "Now c
hoose A1 in such a way that f1=f-A1' has smaller pole order than f." }
}{PARA 0 "" 0 "" {TEXT -1 129 "The highest poles of f are of order 3, \+
so A1' should have pole order 3, so A1 should have pole order 2, these
 poles must be at d." }}{PARA 0 "" 0 "" {TEXT -1 233 "We've seen befor
e that (because multiples of d won't help in the numerator of A1, so w
e can replace the numerator of A1 by its remainder mod d, so we may as
sume that the numerator of A1 has degree at most degree(d,x)-1) we can
 write:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "A1 := add(c[i]*x
^i, i=0..degree(d,x)-1) / d^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A
1G*&,&&%\"cG6#\"\"!\"\"\"*&&F(6#F+F+%\"xGF+F+\"\"\"*$),&*$)F/\"\"#F0F+
F+F+\"\"#F0!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 207 "Now compute \+
f1,  multiply by d^3, take numerator, this should be divisible by d, s
o divide by d and take the remainder, this remainder (hence all its co
efficients) should be zero, compute those coefficients:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "f1 := normal(f-diff(A1,x)): coeffs(
rem(numer(normal(f1*d^3)),d,x),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$
,&!\"%\"\"\"&%\"cG6#F%\"\"%,&F$F%&F'6#\"\"!F$" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 31 "The following are the unknowns:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 38 "var := \{seq(c[i],i=0..degree(d,x)-1)\};" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$varG<$&%\"cG6#\"\"!&F'6#\"\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "S := solve(\{%%\},var);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"SG<$/&%\"cG6#\"\"\"F*/&F(6#\"\"!!
\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "A1 := subs(S,A1); f
1 := normal(f-diff(A1,x)); sqrfree(denom(f1),x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#A1G*&,&!\"\"\"\"\"%\"xGF(\"\"\"*$),&*$)F)\"\"#F*F(F(
F(\"\"#F*!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f1G,$*&,0*$)%\"xG
\"\"(\"\"\"\"\"%*$)F*\"\"&F,!\"%*$)F*F-F,!\"\"*$)F*\"\"$F,!#7*$)F*\"\"
#F,!\"#F*!\"'F4\"\"\"F,*&),&F9F>F>F>\"\"#F,)F*\"\"#F,!\"\"F4" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#7$\"\"\"7$7$,&*$)%\"xG\"\"#\"\"\"F$F$F$F+7$F
*F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "Pole order 2 located at:" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "d := x*(x^2+1);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"dG*&,&*$)%\"xG\"\"#\"\"\"\"\"\"F,F,F,F)F
," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "A2 := add(c[i]*x^i, i=
0..degree(d,x)-1) / d^1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A2G*&,(
&%\"cG6#\"\"!\"\"\"*&&F(6#F+F+%\"xGF+F+*&&F(6#\"\"#F+)F/F3\"\"\"F+F5*&
,&*$F4F5F+F+F+\"\"\"F/\"\"\"!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 70 "f2 := normal(f1-diff(A2,x)): coeffs(rem(numer(normal(
f2*d^2)),d,x),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%,&\"\"\"F$&%\"cG6
#\"\"!F$,&\"\"#F$&F&6#F$!\"#,(F$F$&F&6#F*F-F%\"\"$" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 47 "S := solve(\{%\},\{seq(c[i],i=0..degree(d,x
)-1)\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"SG<%/&%\"cG6#\"\"\"F*/
&F(6#\"\"!!\"\"/&F(6#\"\"#F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 68 "A2 := subs(S,A2); f2 := normal(f1-diff(A2,x)); sqrfree(denom(f2)
,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A2G*&,(!\"\"\"\"\"%\"xGF(*$
)F)\"\"#\"\"\"F'F-*&,&F*F(F(F(\"\"\"F)\"\"\"!\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#f2G,$*&,(*$)%\"xG\"\"%\"\"\"\"\"#*$)F*F-F,!\"%!\"$\"
\"\"F,*&F*\"\"\",&F.F2F2F2\"\"\"!\"\"!\"#" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#7$\"\"\"7$7$,&*$)%\"xG\"\"#\"\"\"F$F$F$F$7$F*F$" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "OK, pole order dropped. Now we hav
e poles of order 1, so we have to compute residues:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 44 "a := numer(f2); b:=denom(f2); bp:=diff(b,
x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG,(*$)%\"xG\"\"%\"\"\"!\"%
*$)F(\"\"#F*\"\")\"\"'\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG
*&,&*$)%\"xG\"\"#\"\"\"\"\"\"F,F,F,F)F," }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#bpG,&*$)%\"xG\"\"#\"\"\"\"\"$\"\"\"F," }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 29 "R := resultant(a-z*bp, b, x);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"RG*&),&!\"'\"\"\"%\"zG\"\"#F+\"\"\",&\"\"'F)F*
!\"\"F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "solve(R,z);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"$F#\"\"'" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 16 "residues := \{%\};" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%)residuesG<$\"\"$\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 48 "A3 := add( r * ln(gcd(a-r*bp, b)), r=residues );" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#A3G,&-%#lnG6#,&*$)%\"xG\"\"#\"\"\"\"\"\"F/F/
\"\"$-F'6#F,\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "f3 := \+
normal(f2 - diff(A3,x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f3G,$%
\"xG!\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 151 "Now the poles of ord
er 1 are also gone, so we're left with a polynomial. Note that we coul
d have found this polynomial right at the beginning by doing:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "quo(numer(f),denom(f),x);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$%\"xG!\"%" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 10 "So we get:" }}{PARA 0 "" 0 "" {TEXT -1 26 "int(f,x) = A
1 + int(f1,x)." }}{PARA 0 "" 0 "" {TEXT -1 27 "int(f1,x) = A2 + int(f2
,x)." }}{PARA 0 "" 0 "" {TEXT -1 27 "int(f2,x) = A3 + int(f3,x)." }}
{PARA 0 "" 0 "" {TEXT -1 58 "And int(f3,x) is easy, integrating -4*x t
hat gives -2*x^2." }}{PARA 0 "" 0 "" {TEXT -1 57 "Hence, combining thi
s, we get: int(f,x) = A1+A2+A3-2*x^2." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "result := A1+A2+A3-2*x^2;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%'resultG,,*&,&!\"\"\"\"\"%\"xGF)\"\"\"*$),&*$)F*\"\"#
F+F)F)F)\"\"#F+!\"\"F)*&,(F(F)F*F)F/F(F+*&F.\"\"\"F*\"\"\"F3F)-%#lnG6#
F.\"\"$-F:6#F*\"\"'F/!\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "Lets
 check this:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "normal( dif
f(result,x) - f );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "OK, we have indeed calculated the \+
antiderivative of f." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 43 "Now do the same for this function in class:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "f := (x^8-3*x^6-13*x^4-3*x^5+x^3-11
*x^2-2)/(x^9+3*x^7+3*x^5+x^3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"
fG*&,0*$)%\"xG\"\")\"\"\"\"\"\"*$)F)\"\"'F+!\"$*$)F)\"\"%F+!#8*$)F)\"
\"&F+F0*$)F)\"\"$F+F,*$)F)\"\"#F+!#6!\"#F,F+,**$)F)\"\"*F+F,*$)F)\"\"(
F+F:F5F:F8F,!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{MARK "33 2 0" 43 }{VIEWOPTS 1 1 0 1 1 1803 }