Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 8, Section 5
Exercise 1a: Integrate 
1
dx using an “old” method. Strap on your helmet!
x  2x  8
2
Now, x 2  2 x  8 = x 2  2 x
8 =
( x  1) 2  3 2 is of the form
Note that
In this case, u =
and a =
, so we’ll use the trig sub u =
. Thus, we have
x 



that and so dx 


 ( x  1) 2  32 

Thus, 
dx
 (x  1)
2
 32

2
1
=

from page 2 of Handout 8.4

=

2

3 sec( ) tan( ) d
=
9 tan 2 ( )
Now, x  1  3 sec( )  sec( ) =
Thus, csc( ) =
Hence, 
1
1
dx =  ln
3
x  2x  8
2
and cot( ) =
x 1
x  2x  8
2

3
x  2x  8
2
+c =
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 8, Section 5
Exercise 1b: Integrate 
It is a fact that
x
Hence,
2
1
dx using an algebraic insight. Sit back and relax.
x  2x  8
2
1
1
1 1
1


= 
.
6 x4
6 x2
x  2x  8
2
1
6( x  4)
Check it:

1
6( x  2)
1
1
dx =
6
 2x  8
1
=
1
1
 x  4 dx  6  x  2 dx =
Exercise 2a: Use the textbook’s PFD method to decompose
Factoring, x 2  2 x  8 =
Thus
2
1
=
x  2x  8
2
Then
1
.
x  2x  8
2
1
=
x  2x  8
2
The “numerator equation” is
The textbook suggests that you substitute in any two (since, in this problem, there are
two unknown coefficients, A and B) numbers for x. The two x-values easiest to use
would be the zeroes of the denominator polynomial.
x = -2
Hence,
1
A
B


=
=
x4 x2
x4 x2
x  2x  8
2
x=4
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 8, Section 5
3
Partial Fraction Decomposition
Partial Fraction Decomposition (PFD) is an algebraic method for rewriting a rational
expression as the sum and/or difference of “simpler” rational expressions.
Exercise 2bob: Use Bob’s PFD method to decompose
1
.
x  2x  8
2
Factoring as in Exercise 2a, x 2  2 x  8 = (x–4)(x+2). Then
Thus
1
A
B

=
.
x4 x2
x  2x  8
2
1
A( x  2)
B( x  4)
=
.

x  2x  8
( x  4)( x  2) ( x  2)( x  4)
2
Moreover, as we saw in Exercise 2a, he “numerator equation” is 1 = A(x + 2) + B(x – 4).
Bob writes
0x + 1 = Ax + 2A + Bx – 4B
Note 1: The method of Partial Fraction Decomposition is practical only for rational
function integrands whose denominators factor “nicely.” For instance, if the denominator
were changed slightly in the last problem from x 2  2 x  8 to x 2  2 x  7 , its
factorization, which is x  2 2  1 x  2 2  1 , would likely be too cumbersome for
the PFD method.



Note 2: In the integrand, if the degree of the numerator is greater than or equal to the
degree of the denominator (i.e., if the integrand is an “improper” fraction), then, as your
first step, divide and rewrite the integrand as a “mixed” fraction.
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 8, Section 5
4
A Case of Distinct Linear and Quadratic Factors
2x3  4x  8
Exercise 3: Determine  2
dx .
x  x x2  4



First, factor the integrand’s denominator as much as possible.
=
A
B
Cx  D
A

 2
=
x x 1 x  4
x
The numerator equation is
Thus,
2x3  4x  8
dx =
x2  x x2  4



+
2x3  4x  8
=
x2  x x2  4


B
( x  1)
+

(Cx  D )
x2  4


2x3  4x  8 =

x
dx + 
x 1
dx + 
x
dx =
x 4
2
2
2
 x dx +  x  1 dx +
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 8, Section 5
5
A Case of Repeated Linear Factors
2x  3
dx .
Exercise 4: Determine  2
x  2x  1
First, factor the integrand’s denominator as much as possible.
So,
2x  3
=
( x  1) 2
A
B

x  1 ( x  1) 2
=
A
( x  1)
+
B
( x  1) 2
The numerator equation is
Hence,
Thus,
2x  3
( x  1) 2
x
2
=
2x  3
dx
 2x  1
x 1
=

( x  1) 2
 x  1 dx
+
2x  3
=
x  2x  1
 ( x  1)
2
dx
2
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 8, Section 5
6
A Case of Repeated Quadratic Factors
x2  x  3
Exercise 5: Determine  4
dx .
x  6x 2  9
x2  x  3
=
x 4  6x 2  9
First, factor the integrand’s denominator as much as possible.
So,
x2  x  3
x
2

3
2
=
Ax  B
Cx  D

2
2
x 3
x2  3

So,
x2  x  3
x
2

3
2
=
( Ax  B )
x 2  3
+

Cx  D
x
2
3

2

x 2  x  3 = ( Ax  B) x 2  3  Cx  D
The numerator equation is
Use Bob’s Method:

=
0 x 3  1x 2  1x  3 =
Ax  B
Cx  D

2
2
x 3
x2  3
x2  x  3
dx =
Thus,  4
x  6x 2  9

x

dx
+
3
2
x
x

2
2
x 3
x 3
=
 x

2
x
2

3
2
dx =

x
dx
3
2
+
 x
x
2
3

2
dx