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\centerline{Mathematics 132 -- Calculus for Physical and Life Sciences 2}
\centerline{Integration by Partial Fractions}
\centerline{February 11 - 12, 2008}
\bigskip
\noindent
The {\it partial fraction} method applies to 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.  
\item{3.}  Set up the partial fractions.  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
$${A_1\over x + a} + {A_2\over (x + a)^2} + \cdots + {A_e\over (x + a)^e}$$
If $(ax^2 + bx + c)^f$ is the highest power of a quadratic with no
real roots that divides $g(x)$, then the partial fractions will include a 
group of terms 
$${B_1x + C_1\over ax^2 + bx + c} + {B_2x + C_2\over (ax^2 + bx + c)^2}+ \cdots + {B_fx + C_f\over (ax^2 + bx + c)^f}.$$
\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
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
$$\eqalign{
\int {x^5 + 4x + 1\over x^4 + 9x^2}\ dx &=
\int x + {-9x^3 + 4x + 1\over x^4 + 9x^2}\ dx\cr
&= \int x  + {4/9\over x} + {1/9\over x^2} + {(-85/9)x + (-1/9) \over x^2 + 9}\ dx\cr
&= {x^2\over 2} + {4\over 9}\ln|x| - {1\over 9}{1\over x} - {85\over 18} \ln(x^2 + 9)
- {1\over 27} \tan^{-1}(x/3) + C.\cr}
$$
\bye 
