SME
Calculus II · Problem type
Partial fractions: two linear factors
Use this for “Evaluate the integral of 1 ÷ ((x − 3)(x + 1)) from 4 to 6.”
Chart
Shaded: ∫ from 2 to 3 of f(x) dx = 0.1567
All of it is above the axis, so the integral is the shaded area
This is an example. Type your own values to start a new problem.
∫
1
(x
−
)(x
−
(
)
)
dx
=
I
A
Top over the first factor
worked out (not typed)
B
Top over the second factor
worked out (not typed)
L₁
Change in ln|x − p|
worked out (not typed)
L₂
Change in ln|x − q|
worked out (not typed)
Formulas
A
=
1
p
−
q
1
3
=
1
1 − (−2)
B
=
1
q
−
p
−
1
3
=
1
−2 − 1
L₁
=
ln|
b
−
p
|
−
ln|
a
−
p
|
0.6931 = ln|3 − 1| − ln|2 − 1|
L₂
=
ln|
b
−
q
|
−
ln|
a
−
q
|
0.2231 = ln|3 − (−2)| − ln|2 − (−2)|
I
=
A
×
L₁
+
B
×
L₂
0.1567
=
1
3
×
0.6931
+
(−
1
3
)
×
0.2231
Assumptions
•
Two different linear factors split as 1 ÷ ((x − p)(x − q)) = A ÷ (x − p) + B ÷ (x − q).
•
Cover-up: multiply by (x − p) and put x = p, so A = 1 ÷ (p − q). In the same way B = 1 ÷ (q − p) = −A.
•
An antiderivative of 1 ÷ (x − r) is ln|x − r|, so the integral is A ln|x − p| + B ln|x − q| + C.
•
The zeros p and q differ, and neither lies from a to b, where the integral would be improper.
Step-by-step
WE KNOW
p = 1, q = −2, a = 2, b = 3
FIND
top over the first factor (A), top over the second factor (B), change in ln|x − p| (L₁), change in ln|x − q| (L₂), value of the integral (I)
Step 1 · Find top over the first factor (A)
Use A = 1 ÷ (p − q)
Cover-up:
multiply
both
sides
by
(x
−
p
),
then
put
x
=
p
.
The
B
term
drops
out
and
1
p
−
q
is
left.
A
=
1
1 − (−2)
A
=
1
3
A
=
1
3
Step 2 · Find top over the second factor (B)
Use B = 1 ÷ (q − p)
Cover
up
(x
−
q
)
the
same
way
and
put
x
=
q
.
B
is
always
−
A
.
B
=
1
−2 − 1
B
=
1
−3
f
(x)
=
1
(x − 1)(x + 2)
=
1
3(x − 1)
−
1
3(x + 2)
B
=
−
1
3
Step 3 · Find change in ln|x − p| (L₁)
Use L₁ = ln|b − p| − ln|a − p|
An
antiderivative
of
1
x −
p
is
ln|x
−
p
|.
Take
its
value
at
a
from
its
value
at
b
.
L₁
=
ln|3
−
1|
−
ln|2
−
1|
L₁
=
ln
2
−
ln
1
L₁
=
0.6931
−
0
L₁
=
0.6931
Step 4 · Find change in ln|x − q| (L₂)
Use L₂ = ln|b − q| − ln|a − q|
The
same
for
the
second
factor:
ln|x
−
q
|
from
a
to
b
.
L₂
=
ln|3
−
(−2)|
−
ln|2
−
(−2)|
L₂
=
ln
5
−
ln
4
L₂
=
1.6094
−
1.3863
L₂
=
0.2231
Step 5 · Find value of the integral (I)
Use I = A × L₁ + B × L₂
Each fraction’s integral is its top times its change in ln. Add them.
I
=
1
3
×
0.6931
+
(−
1
3
)
×
0.2231
∫
1
(x − 1)(x + 2)
dx
=
(
1
3
)(ln|x
−
1|
−
ln|x
+
2|)
+
C
3
∫
2
dx
(
x
− 1)(
x
+ 2)
=
[(
1
3
)(ln|x
−
1|
−
ln|x
+
2|)]
from
2
to
3
=
0.1567
I
=
0.1567
Check
Put the numbers back in. Both sides should match.
1
3
=
1
1 − (−2)
✓
−
1
3
=
1
−2 − 1
✓
0.6931 = ln|3 − 1| − ln|2 − 1| ✓
0.2231 = ln|3 − (−2)| − ln|2 − (−2)| ✓
0.1567
=
1
3
×
0.6931
+
(−
1
3
)
×
0.2231
✓
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