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\centerline{Mathematics 136 -- Calculus 2}
\centerline{Summary of Partial Fractions}
\centerline{February 17 and 18, 2005}
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Every rational function (quotient of polynomials) can 
be written as {\it a polynomial plus a sum of one or more terms of the following 
forms}:
$${Ax+B\over (x^2+2bx+c)^k},\qquad {C\over (x-a)^k}.$$
A rational function expressed this way is said to be {\it decomposed
into partial fractions}.  The process of finding this decomposition
is as follows:  Given a rational function ${f(x) \over g(x)}$,
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\item{1.}  First, if the degree of $g$ is larger already, just
proceed to step 2 with ${f(x)\over g(x)}$.  
Otherwise, if the degree of $f(x)$ is greater than or equal to the 
degree of $g(x)$, {\it divide} $g(x)$ into $f(x)$ using polynomial
long division and write $f(x) = q(x)g(x) + r(x)$ for some
quotient $q(x)$ and remainder $r(x)$.  Then 
$${f(x)\over g(x)} = q(x) + {r(x)\over g(x)}$$
and the degree of $r(x)$ is less than the degree of $g(x)$.  Continue 
to step 2 with ${r(x)\over g(x)}$.  
\item{2.}  {\it Factor} $g(x)$ completely into a product of 
powers of linear polynomials and powers of irreducible quadratic
polynomials.  (The fact that this can always be done is one
form of the so-called ``Fundamental Theorem of Algebra''.  The
famous mathematician and physicist Carl Friedrich Gauss gave the first
complete proof of this result in 1799.)
\item{3.}  Assemble the partial fractions:  For each $(x-a)^m$ appearing 
in the factorization of $g(x)$, include a {\it group of terms}
$${C_1\over (x-a)} + {C_2\over (x-a)^2} + \cdots + {C_m\over (x-a)^m}$$
For each $(x^2+2bx+c)^n$ appearing 
in the factorization of $g(x)$, include a {\it group of terms}:
$${A_1x+B_1\over x^2+2bx+c} + {A_2x+B_2\over (x^2+2bx+c)^2} + \cdots + {A_n x+B_n\over (x^2+2bx+c)^n}$$
\item{4.}  Set the rational function from step 1 (either the original
$f/g$ or $r/g$ as appropriate) equal to the sum of the partial fractions,
clear denominators, and solve for the coefficients.  
This last step can be done either by substituting well-chosen $x$-values,
or by equating coefficients of like powers of $x$ on both sides
and solving the resulting system of equations.
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\noindent
An interesting consequence of the partial fraction method is that it
shows, {\it in principle}, that 
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\proclaim Theorem. Every rational function has elementary antiderivatives.\par
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\noindent
 We add the ``in principle'' here because
determining the factorization of $g(x)$ will not be simple in
some cases, and the methods that allow 
us to integrate the partial fractions with higher powers
of quadratic polynomials in the denominator involve trig substitutions that
can get ``hairy.''  

In any case, here's an example showing most of the different things that can 
happen.  Say we want to decompose 
$${x^4+1\over x^4 + x^2}$$ 
into partial fractions.  First, the degree of the top is $4$ and the 
degree of the bottom is $4$ also, so we begin by dividing the 
bottom into the top:
$$x^4 + 1 = 1(x^4 + x^2) + (- x^2 + 1)$$
This means 
$${x^4+1\over x^4 + x^2} = 1 + {- x^2 + 1\over x^4 + x^2}$$
We continue to step 2 with the rational function 
$${-x^2 + 1\over x^4 + x^2}$$
The factorization of the denominator is 
$$x^4 + x^2 = x^2(x^2 + 1)$$
Hence our general recipe for assembling the partial fractions says:
$${-x^2 + 1\over x^4 + x^2}
= {C_1 \over x} + {C_2 \over x^2} + {A x + B \over x^2 + 1}$$
So after clearing denominators and collecting terms,
$$-x^2 + 1 = C_1x(x^2+1) + C_2(x^2+1) + (A x + B)x^2
=(A + C_1)x^3 + (B+C_2)x^2+C_1 x + C_2$$
Equating coefficients of like powers of $x$,
$$C_2 = 1, C_1 = 0, B+C_2 = -1, A + C_1 = 0,$$
so 
$$C_2 = 1, C_1 = 0, B = -2, A = 0.$$
In other words, 
$${x^4+1\over x^4 + x^2} = 1 + {-x^2 + 1\over x^4 + x^2} = 1 + {1 \over x^2} + {-2 \over x^2 + 1}$$
From this we can integrate:
$$\int {x^4+1\over x^4 + x^2}\ dx
 = \int 1 + {1 \over x^2} + {-2 \over x^2 + 1}\ dx
 = x + {-1\over x}  - 2\arctan(x) + C$$
\bye
