Integration via Partial Fractions: Example 2
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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.
How do we proceed?
We can determine
A
and
B
by combining the terms on the right hand side:
What is the least common denominator for them?
(
x
– 5)*(
x
+ 3)
Combine the fractions.
Distribute the multiplication on the right-hand side.
We get
Collect like terms in
x
.
What do we do next?
We equate coefficients of like powers of
x
.
Do that.
For
Equation 1
For
Equation 2
We need to solve these simultaneously. The
Substitution Method
looks like an easy way here. Solve equation 1 for
B
in terms of
A
.
Equation 3
Substitute this into equation 2.
Solve for
A
.
Use this in equation 3 to find
B
.
Use these results to rewrite
F
in terms of partial fractions.
Substitute in the original integral.
We get
To see what type of result will we get from these terms substitute
u
=
x
– 5 in the first term, and
v
=
x
+ 3 in the second.
We get
What type of result do we get from each term?
We get a
natural log form
.
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