Download Solution

Math 2300: Calculus I
Partial Fractions
We can integrate some rational rational functions using a u-substitution or trigonometric substitution, but this method does not always work. We now will learn a method that allows us to express
ANY rational function as a sum of functions that can be integrated using old methods. That is,
1
1
1
we cannot integrate 2
as-is, but it is equivalent to
− , each term of which, we can
x −x
x−1
x
integrate.
Z
x−7
1. Goal: Compute
dx.
(x + 1)(x − 3)
Z
1
(a)
dx =
x+1
Z
1
dx = ln |x + 1| + C
Solution:
x+1
(b)
2
1
−
=
x+1 x−3
1
2(x − 3) − (x + 1)
x−7
2
−
=
=
x+1 x−3
(x + 1)(x − 3)
(x + 1)(x − 3)
Solution:
Z
(c)
x−7
dx =
(x + 1)(x − 3)
Z
Solution:
Z
2. Goal: Compute
(a)
Z
2
1
−
dx = 2 ln |x + 1| − ln |x − 3| + C
x+1 x−3
10x − 31
dx.
(x − 1)(x − 4)
10x − 31
3
7
=
+
. Verify your answer.
(x − 1)(x − 4)
x−4
x−1
Z
(b)
x−7
dx =
(x + 1)(x − 3)
10x − 31
dx =
(x − 1)(x − 4)
Z
3
dx +
x−4
Z
7
dx =
x−1
Solution:
3 ln |x − 4| + 7 ln |x − 1| + C
1
Math 2300: Calculus I
Z
3. Goal: Compute
(a)
Partial Fractions
x + 14
dx.
(x + 5)(x + 2)
?
?
x + 14
=
+
(x + 5)(x + 2)
x+5
x+2
There is no indicator of what the numerators should be, so there is work to be done to
find them. If we let the numerators be variables, we can use algebra to solve. That is, we
want to find constants A and B that make equation (1) below true for all x 6= −5, −2,
which are the same constants making equation (2) below true for ALL x.
x + 14
A
B
=
+
(x + 5)(x + 2)
x+5 x+2
x + 14 = A(x + 2) + B(x + 5)
1
x + 14
=
A+B
x+
(1)
(2)
2A + 5B
(3)
Note: Two polynomials are equal if corresponding coefficients are equal. For linear
functions, this means that ax + b = cx + d for all x exactly when a = c and b = d )
Alternatively, you can evaluate equation (2) for various x0 s to get equations relating A
and B. Some values of x will be more helpful than others.
(
1
14
=
=
A+B
2A + 5B
Continue solving for the constants A and B.
Solution:
Z
(b)
(
A+B
2A + 5B
=1
⇒
= 14
(
A+B
3B
=1
⇒
= 12
(
A
B
= −3
=4
x + 14
dx =
(x + 5)(x + 2)
Solution:
Z
x + 14
dx =
(x + 5)(x + 2)
Z
−3
4
+
dx = −3 ln |x + 5| + 4 ln |x + 2| + C
x+5 x+2
2
Math 2300: Calculus I
Z
4.
Partial Fractions
x + 15
dx =
(3x − 4)(x + 1)
Solution:
x + 15
A
B
=
+
(3x − 4)(x + 1) 3x − 4 x + 1
x + 15= A(x + 1) + B(3x − 4)
x = −1 :
x=
Z
4
:
3
x + 15
dx =
(3x − 4)(x + 1)
Z
14 = A · 0 + B · (−7) ⇒ B = −2
49
7
=A· +B·0⇒A=7
3
3
7
−2
7
+
dx = ln |3x − 4| − 2 ln |x + 1| + C
3x − 4 x + 1
3
3
Math 2300: Calculus I
Z
5. Goal:
Partial Fractions
5x − 2
dx
(x + 3)2
Here, there are not two different linear factors in the denominator. This cannot be expressed
in the form:
5x − 2
5x − 2
A
B
A+B
=
6=
+
=
(x + 3)2
(x + 3)(x + 3)
x+3 x+3
x+3
However, it can be expressed in the form:
A
5x − 2
B
=
+
2
(x + 3)
x + 3 (x + 3)2
(a) Find constants A and B that make the above equation true for all x 6= −3.
Solution:
5x − 2
A
B
=
+
(x + 3)2 x + 3 (x + 3)2
5x − 2= A(x + 3) + B
x = −3 :
x=0:
Z
(b) Compute
−17 = A · 0 + B ⇒ B = −17
−2 = A · 3 + (−17) ⇒ A = 5
5x − 2
dx
(x + 3)2
Solution:
Z
5x − 2
dx =
(x + 3)2
Z
5
−17
17
+
dx = 5 ln |x + 3| +
+C
x + 3 (x + 3)2
x+3
4