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\centerline{Mathematics 136 -- AP Calculus}
\centerline{Integration by Partial Fractions}
\centerline{Fall 2009}
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\noindent
The {\it partial fractions} method applies to (in principle, {\it all}) rational functions
$$h(x) = {f(x)\over g(x)} = {a_n x^n + \cdots + a_1 x + a_0\over b_m x^m + \cdots + b_1 x + b_0}$$
(or to functions that can be brought to this form by a preliminary substitution).
The steps involved are:
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\item{1.}  If $n \ge m$, first {\it divide} $g(x)$ into $f(x)$ using polynomial division
to write $f(x) = q(x)g(x) + r(x)$, where the degree of $r(x)$ is less than $m$.
This yields 
$${f(x)\over g(x)} = q(x) + {r(x)\over g(x)}$$
and 
$$\int {f(x)\over g(x)}\ dx = \int q(x)\ dx + \int {r(x)\over g(x)}\ dx$$
\item{2.}  Now, assuming we have a rational function where the degree of the 
numerator is strictly less than the degree of the denominator, {\it factor} the 
denominator completely into linear and quadratic factors.  The quadratic 
factors will have no real roots when they cannot be factored further.  
(It is a theorem that any polynomial with real coefficients can be factored
in this way.  The ``bottleneck'' in applying this method in very hard examples, 
though, is often finding the appropriate factorization.  {\it Needless to say,
we will not be doing any of those very hard examples(!)} 
\item{3.}  Set up the partial fractions. 
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\itemitem{a)} If $(x + a)^e$ is the highest power 
of a linear polynomial that divides $g(x)$, then the partial fractions will 
include a group of terms
$${A_1\over x + a} + {A_2\over (x + a)^2} + \cdots + {A_e\over (x + a)^e}$$
\itemitem{b)}
If $(ax^2 + bx + c)$ is a quadratic with no
real roots that divides $g(x)$, then the partial fractions will include a term 
$${Bx + C\over ax^2 + bx +c}.$$
Note the {\it linear polynomial on the top}.  The reason you need both the
$Bx$ and the $C$ will become apparent by doing some examples.  It is not 
possible to ``match'' every possible numerator unless both the $Bx$ and the $C$ are 
included.  {\it Note:}
We will only be doing examples where quadratic factors appear to the first
power in the factorization of the denominator.  There is a more general
form of partial fractions that applies if you have something like 
$(x^2 + 6x + 10)^3$ in the factorization, though.  You need a sum of 
partial fractions in that case, something like in step 3 above.   
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\item{4.}  Solve for the coefficients in the partial fractions.  This can be done 
by clearing denominators, then substituting in $x$-values and/or equating coefficients
on both sides of the resulting equation.
\item{5.}  Integrate the partial fractions.  
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\noindent
{\it Worked Example}
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Here is an example illustrating many of these steps.  Suppose we need to integrate
$$\int {x^5 + 4x + 1\over x^4 + 9x^2}\ dx.$$
The degree of the top is larger, so we divide first:
$$x^5 + 4x + 1 = x (x^4 + 9x^2) + (-9x^3 + 4x + 1)$$
so 
$${x^5 + 4x + 1\over x^4 + 9x^2} = x + {-9x^3 + 4x + 1\over x^4 + 9x^2}$$
The denominator factors as $x^4 + 9x^2 = x^2(x^2 + 9)$.  So the partial fractions
are 
$$ {-9x^3 + 4x + 1\over x^4 + 9x^2} = {A\over x} + {B\over x^2} + {Cx + D\over x^2 + 9}.$$
Clearing denominators,
$$-9x^3 + 4x + 1 = Ax(x^2 + 9) + B(x^2 + 9) + (Cx + D)x^2.$$
Substituting $x = 0$ we see $1 = 9B$ so $B = 1/9$.  From the coefficient of $x$
we see $4 = 9A$, so $A = 4/9$.  Then from the coefficients of $x^3$, 
$A + C = -9$, so $C = -9 - 4/9 = -85/9$ and finally from the coefficient of $x^2$, 
$0 = B + D$, so $D = -1/9$.  This gives
$$\eqalign{
\int {x^5 + 4x + 1\over x^4 + 9x^2}\ dx &=
\int x + {-9x^3 + 4x + 1\over x^4 + 9x^2}\ dx\cr
&= \int x  + {4/9\over x} + {1/9\over x^2} + {(-85/9)x + (-1/9) \over x^2 + 9}\ dx\cr
&= {x^2\over 2} + {4\over 9}\ln|x| - {1\over 9}{1\over x} - {85\over 18} \ln(x^2 + 9)
- {1\over 27} \tan^{-1}(x/3) + C.\cr}
$$
A final comment -- note that if the $Cx$ term were not present, then there would
be no way to ``match'' the numerator $-9x^3 + 4x + 1$ when we add up the partial fractions.
\bye 
