M408C - Notes - Integrals for Final - Contents 2 3 4 5 6 7...

Contents
2 Exam Review: Long Division and Complete-the-Square Example
3 Finishing Off the Complete-the-Square Example
4 The Integration by Parts example
5 Which also involves trig sub
6 Trig Integrals
7 A Third Trig Integral
8 Trig Sub
9 The Last Trig Subbie

Exam Review: Long Division and Complete-the-Square Example
Here it is, the exam review sheet. The topics to be covered are:
a) Long division; completing the square
b) Integration by parts
c) Trig Integrals
d) Trig substitution
e) Partial fractions
A) Example
R
x
3
+1
x 2 +2 x +3
dx .
When you have a quotient of polynomials, it’ll be either a complete
the square problem or a partial fractions, but in either case you’ll
need to do a long division first. Then
Z
x
3
+ 1
x 2 + 2 x + 3
dx =
Z
( x - 2) +
x + 7
x 2 + 2 x + 3
¶
dx
What happens next depends whether the denominator factors or
not: wherether b
2
- 4 ac ≥ 0 or b
2
- 4 ac < 0. But a = 1 , b = 2 , c = 3,
and b
2
- 4 ac = 2
2
- 4(1)(3) < 0 so the denominator does not factor,
and this is a complete-the-squares problem.
Completing the square: x
2
+2 x +3 = ( x + e )
2
+ f = x
2
+2 ex +( e
2
+ f ),
so ya got 2 e = 2 and e
2
+ f = 3. Then e = 1 and 1+ f = 3 or f = 2.
In all, x
2
+ 2 x + 3 = ( x + 1)
2
+ 2 and
Z
x
3
+ 1
x
2
+ 2 x + 3
dx =
Z
( x - 2) +
x + 7
x
2
+ 2 x + 3
¶
dx
=
Z
( x - 2) dx +
Z
x + 7
( x + 1) 2 + 2
dx
=
x
2
2
- 2 x
¶
+
Z
x + 7
( x + 1) 2 + 2
dx
Now do a u -substitution in the integral: u = x +1; du = dx ; x +7 =
u + 6. Then

Finishing Off the Complete-the-Square Example
Z
x
3
+ 1
x
2
+ 2