Integration via Partial Fractions: Example 5
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Solve
, using partial fractions.
Factors in the denominator, as we have here, indicate that we can use partial fractions.
Using just the integrand above, set up the partial fraction equivalent. See this excellent
reference site
for techniques.
Equation 1
How do we proceed?
We can determine
A
,
B
,
C
, and
D
by combining the terms on the right hand side:
What is the least common denominator for them?
x
2
* (
x
– 7) * (
x
– 1)
Combine the fractions so that we can equate the numerators.
Using the least common denominator, we get
Equation 2
How can we solve for
A
,
B
,
C
, and
D
?
There are two methods:
a) We can take advantage of the fact that this equation must hold for all values of
x
. Accordingly, we can choose simplifying values of
x
, like
x
= 0,
x
= 1, and
x
= 7.
b) Alternatively, we can equate coefficients of like terms and solve four equations simultaneously.
We’ll do both as a check, and start with (a). Substitute
x
= 0 in the numerator on both sides of equation 2.
Substitute
x
= 1 in the numerator on both sides of equation 2.
Substitute
x
= 7 in the numerator on both sides of equation 2.
Substitute
x
= 2 in the numerator on both sides of equation 2.
Substitute in this equation the values of
B
,
C
, and
D
just obtained.
Solve for
A
.
Now we can check these four values by using method (b). First distribute the multiplication in the numerators of equation 2.
Collect like powers of
x
on the right-hand side.
Equate coefficients of like powers.
Equation 3
Equation 4
Equation 5
Equation 6
We have 4 equations in the 4 unknowns:
A
,
B
,
C
, and
D
. How can we solve for them?
We can use the
Substitution and Elimination Methods
.
How should we proceed?
Solve for
B
in equation 6, and substitute in equation 5 to get
A
. After that, use those values to simplify equation 3 and 4 and use elimination.
Find
B
.
From equation 6,
B
= 1
Find
A
.
From equation 5 with
B
= 1 we get
Use these results to simplify equations 3 and 4.
We get
Add these resulting two equations to eliminate
C
.
We get
Find
C
.
Using one of the equations above, we get
In summary, we have agreement between our two methods.
Use these results to rewrite equation 1 above to express
F
in partial fractions.
Substitute in the original integral.
We get
To see what type of result we will get from these terms, convert the second term to a negative power, substitute
u
=
x
–7 in the third term and
v
=
x
–1 in the fourth.
We get
What type of result do we get from each term?
We get a
natural log form
from the first, third and fourth terms. We can use the
Power Rule of Integration
on the second term.
Do the integration.
Are we done?
No.
What next?
We convert back to expressions in
x
.
Do that.
How can we check this result?
We can take the derivative,
.
What do we then compare it to?
We compare it to the integrand of the original problem.
Determine this derivative.
Combine these fractions.
We get
which checks.
The end. If you found this helpful and would recommend that I create more pages like this one, please let me know:
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General Contents
Detailed Contents
Index