prob4sol - Math 151 Spring 208 Geveci Problem Set # 4...

Math 151 Spring 208 Geveci
Problem Set # 4
Integrals of Rational Functions
Solutions
1. The expression 4 x
2
+ 8 x + 5 is irreducible since 8
2
− 4 (5) (4) = − 16 < 0 . We complete
the square:
4 x
2
+ 8 x + 5 = 4
¡
x
2
+ 2 x
¢
+ 5 = 4 ( x + 1)
2
− 4 + 5 = 4 ( x + 1)
2
+ 1 .
Therefore,
Z
1
4 x 2 + 8 x + 5
dx =
Z
1
4 ( x + 1)
2
+ 1
dx =
Z
1
(2 ( x + 1))
2
+ 1
dx.
We set u = 2 ( x + 1) so that du = 2 dx . Thus,
Z
1
(2 ( x + 1))
2
+ 1
dx =
1
2
Z
1
u 2 + 1
du =
1
2
arctan ( u ) =
1
2
arctan (2 x + 2) .
2. The expression x
2
+ 2 x + 5 is irreducible since 2
2
− 4 (5) = − 16 < 0 . We complete the
square:
x
2
+ 2 x + 5 = ( x + 1)
2
+ 4 .
Thus,
Z
x
x
2
+ 2 x + 5
dx =
Z
x
( x + 1)
2
+ 4
dx
=
Z
x + 1
( x + 1)
2
+ 4
dx −
Z
1
( x + 1)
2
+ 4
dx
=
1
2
ln
¡
x
2
+ 2 x + 5
¢
−
1
2
arctan
μ
x + 1
2
¶
3.
a)
− 3 x + 16
x
2
+ x − 12
=
− 3 x + 16
( x + 4) ( x − 3)
=
A
x + 4
+
B
x − 3
⇔
− 3 x + 16 = A ( x − 3) + B ( x + 4) .
Set x = − 4 :
28 = − 7 A ⇒ A = − 4 .
Set x = 3 :
7 = 7 B ⇒ B = 1 .
Thus,
− 3 x + 16
x 2 + x − 12
= −
4
x + 4
+
1
x − 3
1

b)
Z
− 3 x + 16
x 2 + x − 12
dx = − 4
Z
1
x + 4
dx +
Z
1
x − 3
dx
= − 4 ln ( | x + 4 | ) + ln ( | x − 3 | ) .
4.
a)
x + 7
( x − 2) ( x − 5)
=
A
x − 2
+
B
x − 5
⇔
x + 7 = A ( x − 5) + B ( x − 2)
Set x = 2 :
9 = − 3 A ⇒ A = − 3 .
Set x = 5 :
12 = 3 B ⇒ B = 4 .
Thus,
x + 7
( x − 2) ( x − 5)
= −
3
x − 2
+
4
x − 5
b)
Z
x + 7
( x − 2) ( x − 5)
dx = − 3
Z
1
x − 2
dx + 4
Z
1
x − 5
dx
= − 3 ln ( | x − 2 | ) + 4 ln ( | x − 5 | ) .