



When
we want to find an integral like this one ...












we
have a big problem. 


Up
to now, our only trick with stuff like this was the old ... 











routine. 


The
odds of that working on something this big are not so good. 


In
this example, we have a bigger exponent in the numerator 


than
anything in the denominator, 


so
there's NO chance of it working. 





A
fraction like this one, with a larger power of X in the numerator 


than
any we have in the denominator, 


is
called an improper fraction. 





If
the biggest exponent in a fraction is in the denominator, 


it
is called a proper fraction. 


They're
easier to deal with, calculus wise anyway ... 





WELL
WE DON'T HAVE A PROPER FRACTION, 


WHAT
DO WE DO NOW? 





Just
turn it into a proper fraction of course! 





Oh
yeah? How do we do that? 





We
use our good buddy, long division. 


And
with what we've got here, we're talkin' LONG division. 











So
that means ... 











Most
of this new thing is easy to deal with. 


We'll
go as far as we can. 


We
get ... 











We
still have a tough part, but now it's a lot smaller. 


We
still can't use the lnX trick (we got close!). 


We
don't have a "+ C" part yet, 


we
can say it's still hiding inside the last integral part. 





Now
we deal with the last part by doing a strange thing. 


We
factor the denominator! 











What
we want to do is turn this last term into two fractions. 


Each
fraction will have a constant in the numerator, 


and
one of our two factors in the denominator. 





We
want to have ... 











Once
upon a time (hopefully) you learned how to add two fractions 


by
finding a common denominator. 


Here
we are doing just the opposite. 


We're
ripping a single fraction apart into two pieces that, 


we
hope, will be easier to deal with. 





The
next step to solve this is to clear the denominators. 


To
do that, multiply both sides by (X  1)(X  2). 


You
get ... 





2X
+ 12 = A(X  2) + B(X  1)






Multiply
out the right side ... 





2X
+ 12 = AX  2A + BX  B






Rearrange
a bit ... 





2X
+ 12 = AX + BX  2A  B






Now
factor the right side ... 





2X + 12 = (A +
B)X  2A  B 





Now
here's a great trick. 


Since
the left side of the equation is equal to the right. 


(Hey,
that's what an equation does!) 


That
means the coefficient on the X term on the left side (the 2), 


is
equal to the coefficient on the X term on the right side (the A +
B)! 


So
... 


2 = A + B






and,
and, and ... 


By
the same logic, the number on the left (12) 


MUST
be equal to the numbers on the right! 


So
... 


12
= 2A  B






So
now we have two equations with two unknowns. 


That
means we can solve for them ... 











We
can solve for B by substitution ... 











So
A = 14, and B = 16. 


That
means our remaining integral part (remember the problem?) 


is
... 











so
cleaning up these two a bit ... 











And
now we finally get to use the "ln" thing! 











Remember,
something like ln5 doesn't mean anything, 


so
we use absolute value notation to make sure we get 


ln(something
positive). 


Tacking
this on to the part we already solved, we have ... 











We
might have expected this one to have a long and messy answer! 





Review
Time: 





When
we have an improper fraction, 


we
divide the numerator by the denominator 


to
get some terms all by themselves 


AND
a proper fraction. 





And
now we also have a way to integrate a proper fraction 


even
when it isn't a du/u setup to begin with. 











There
is another way that you can use to figure out A and B. 





We'll
use the proper fraction part from the last example 


so
that you see we get the same answers. 











First,
factor the denominator of the fraction ... 











This
time, we cover up one of the factors of the denominator ... 











The
X value that makes the covered up factor equal to zero is 1. 


So
we solve the rest of the fraction for X = 1 ... 











This
is the numerator of one of our integrals. 


The
factor that we covered up, is the denominator. 


So
that part of the integral is ... 











To
find the other piece, cover the other denominator factor 


and
solve the rest for the X value that makes THAT factor 


equal
to zero (X = 2) ... 











So
the other integral term is ... 











You
can check these with what we got before if you want. 


It's
the same integrals. 





Let's
do another one ... 





Example: 











The
first thing that we need to do, 


is
check to see if this is a proper fraction or not. 


The
numerator is degree 2 (highest exponent term is 3X
^{2}). 


The
denominator is degree 3 (highest exponent term is X
^{3}). 


That
means we have a proper fraction. 


We
get a big break, because the denominator has already been
factored. 





Simplify
this thing using the original method ... 











Multiply
both sides by (X  1)(X  2)(X  3). 


We
get ... 





3X^{2}
 5X + 2 = A(X2)(X3) + B(X1)(X3) + C(X1)(X2)






Multiply
it all out ... 





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






Rearrange
the right side ... 





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






So
from the coefficients we have ... 





A + B + C = 3 






5A  4B  3C =  5






6A + 3B^{ }+
2C = 2 





Solve
these any way you like, you get ... 





A =
0 B =  4 C = 7 





So
if we can remember which letter 


was
over which part of the denominator, 


we
have ... 











Using
the "cover up" method, we would do this ... 





Step
1, find A ... 











Step
2, find B ... 











Step
3, find C ... 











These
are the same answers we got using the other method. 


Most
people find this method easier and faster to use, 


but
use either one you like. 


Neither
one is any more correct than the other. 


They
are both just algebra tricks. 





copyright 2005 Bruce Kirkpatrick 
