Integration via Partial Fractions: Example 1
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Solve
using partial fractions.
Since we need factors in the denominator, we must factor the denominator if possible.
Using just the fraction above, try to factor the denominator.
Set up the partial fraction equivalent. See this excellent
reference site
for techniques.
We can determine
A
and
B
by combining the terms on the right hand side:
How does this help?
We know that this can be satisfied by all values of
x
, and that the numerators alone must be equal for all values of
x
.
What are some convenient values of
x
?
Any values of
x
which simplify the equation. In particular, we can use convenient values of
x
: –1 and 0.
Try
in
.
We get
, or
Now try
and use this value of
B
= –7.
We get
, or
How do we use these values?
We can substitute these values into the original fraction,
F
.
Do that.
Substitute in the original integral.
We get
Rewrite the second term so that the power is in the numerator.
To see what type of result will we get from these terms substitute u = x+1 in the first term, and v = x+1 in the second.
We get
What type of result do we get from each term?
We get a
natural log form
in the first term, and a
Power Rule of Integration
result in the second.
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