



When
we need to find the integral of a fraction



where
one of the factors in the denominator is repeated, 


there
is a complication. 











The
denominators we wind up with might be a surprise. 


They
are (X + 3), (X  2) and ... 


(X
 2) ^{2}. 











Before
we go on, there is some bad news. 


The
"cover up the factor" trick won't work 


when
two or more of the denominator factors are the same. 


That
means we are left with the standard method. 


Step
one of the standard method is to get rid of the denominator. 


To
do it, multiply both sides by (X  2)
^{2}(X + 3). 


We
get ... 





5X
^{2}  9X  22 = (X 
2)(X + 3)A + (X + 3)B + (X  2)
^{2}C 





Multiply
out the right side ... 





5X
^{2}  9X  22 = AX
^{2} + AX  6A + BX +
3B + CX ^{2}
 4CX + 4C 





Gathering
and factoring ... 





5X
^{2}  9X  22 = (A +
C)X ^{2}
+ (A + B  4C)X + ( 6A + 3B + 4C) 





So
... 





A 
+ 


C 
= 
5 







A 
+ 
B 
 
4C 
= 

9 








6A 
+ 
3B 
+ 
4C 
= 

22 






Solve
these any way you like, 


you
get ... 





A =
3 B = 
4 C = 2 





So
we have ... 











Part
1: 








Part
2: 








Part
3: 








Putting
it all together, we get ... 

















You
might have a fraction 


where
the denominator can't be factored out 


to
linear terms (that means no exponents on the X's). 


We
might have ... 











The
first thing to see is that the degree of the numerator (2), 


is
less than the degree of the denominator (3). 


That
means that this is a proper fraction. 


If
it wasn't, we would multiply out the denominator 


and
do long division on this puppy. 


Since
this is a proper fraction already, long division won't help. 


We
start in with the A & B stuff. 


This
time, because we have that non linear term in the denominator, 


the
terms break apart like this ... 











Obviously,
the big deal here is the AX + B in the numerator. 


In
the A & B terms we have had before this, 


the
exponent in the denominator has always been 1. 


We
did not have any X's in the numerators. 


Another
way to say that, is that the exponent on the X in the numerator 


was
ONE LESS in the numerator than in the denominator. 


That
is, X to the zero power. 


What
we have here then is the same deal. 


The
exponent on the X in the numerator is one less 


than
the exponent on the X in the denominator. 





So
much for the explanation, on with the problem! 





The
first step is to multiply both sides to get rid of the denominator. 


That
means we multiply both sides by (X
^{2} + 1)(X  3). 





3X
^{2}  7X  16 = (AX +
B)(X + 3) + (X ^{2}
+ 1)(C) 





Multiply
out the right side ... 





3X
^{2}  7X  16 = AX
^{2}  3AX + BX  3B +
CX ^{2}
+ C 





Rearrange
and refactor ... 





3X
^{2}  7X  16 = (A +
C)X ^{2}
+ ( 3A + B)X +( 3B + C) 





So
we have ... 





A 
+ 


C 
= 
3 








3A 
+ 
B 


= 

7 










3B 
+ 
C 
= 

16 






Solving
these we get ... 





A =
4
B = 5
C = 1






That
makes our integrals ... 











Let's
look at the first integral. 


Split
it into two pieces ... 











The
first part is easy ... 











Since
X ^{2}
+ 1 can't be negative, we can drop the absolute value lines. 











The
second part looks like a trig substitution problem. 


Remember
them? 

















and
from tan q
= X we get q
= tan ^{1}X, 


so
... 











That's
two thirds of the problem done ... 











For
the last part, we substitute: u = X  3, du = dX: 











And
we're done! 





Wow!
What a trip! The original integral was ... 











and
our answer is ... 











Don't
get too impressed with all this substitution 


and
reduction formula stuff we've been doing 


for
the last few pages. 


There
are lots of integrals that can't be solved 


with
any of these tricks. 


Today,
we can grind some of them through computers. 





Long
before there were computers, though, 


there
were Brook Taylor and Colin McLaurin ... 





copyright 2005 Bruce Kirkpatrick 
