Math 11B
Spring 2008
Handout #11
Copyright by Hongyun Wang, UCSC
How to calculate
P( x)
Q( x ) dx ?
Q( x ) = x + b
Case 1:
Long division:
r
P( x)
= R( x ) +
x+b
x +b
P ( x)
x + b dx = R ( x ) dx + r log x + b + c
Q( x ) = x 2 + bx + c
Case 2:
Long division:
rx + s
P( x)
= R( x ) + 2
x + bx + c
x + bx + c
2
x
2
P ( x)
rx + s
dx = R ( x ) dx + 2
dx
+ bx + c
x + bx + c
Therefore, we only need to find how to calculate
x
2
Case 2A: b 4c > 0
Factor Q( x )
Step 1:
x 2 + bx + c = ( x x1 ) ( x x2 )
Step 2:
Partial fraction decomposition:
A
B
rx + s
=
+
( x x1)( x x 2) ( x x1) ( x x 2)
x
2
rx + s
1
1
dx = A dx + B dx
+ bx + c
( x x1 )
( x x2 )
= A log x x1 + B log x x2 + c
-1-
2
rx + s
dx .
+ bx + c
Math 11B
2
Case 2B: b 4c = 0
Factor Q( x )
Step 1:
x 2 + bx + c = ( x x1 )
Step 2:
Partial fraction decomposition:
rx + s
( x x1 )
x
2
2
2
=
A
B
+
( x x1 ) ( x x1 )2
rx + s
1
1
dx = A dx + B dx
+ bx + c
( x x1 )
( x x1 )2
= A log x x1 B
1
+c
x x2
2
Case 2C: b 4c < 0
Step 1:
Complete the square
b
4c b 2
x + bx + c = x + +
4
2
2
2
2
b
= x + +2 ,
2
Step 2:
Use substitution u = x +
rx + s
2
b
x + + 2
2 =
4c b 2
4
b
2
b
rb
r u + s
ru + s 2
2
dx = du = du
2
2
2
2
u +
u +
2
= r
=
1
u
rb
du + s 2
du
2
u +
2 u +2
2
(
r
log u 2 + 2
2
)
rb
s 2
u
+
tan 1 + c
-2-
Math 11B
rb
b
s 2
x
+
r
b
2
2 +c
= log x + + 2 +
tan 1 2
2
Here we have used integration formulas:
x
x
2
2
1
x
2
2
log( x + ) + c
2 dx =
2
+
1
1
1 x tan
2 dx =
+ c
+
-3-