Integration of Log: Example 6
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We wish to find
Equation 1
Can we do this directly?
No.
Why?
Because we don't know of a function for which the derivative is
Isn't the integral equal to
?
No, because we would need a factor of
instead of just dx in the integral.
How can we solve this problem?
We can try Integration by Parts.
Set that up in general.
Equation 2
How do we choose the parts?
Usually we want the most complicated part of the integrand to be
so that
will be simpler. The rest of the integrand will be
.
Express
Express
Express
.
Express
Substitute these results in Equations 1 and 2.
Simplify.
Equation 3
Is this simpler than Equation 1?
Yes, it contains only the first power of the log in the integral.
Can we do this remaining integral with the Power Rule?
No.
Why?
A factor of
is needed.
How can we do it?
We can again use Integration by Parts. Let's use a subscript "2" for this, and only work on the integral. We'll take care of the negative 2 factor later.
Set up Integration by Parts for this remaining integral.
What shall we use for
?
Again, let
represent the more complicated part.
Which part is that here?
The logarithmic factor.
Define
Express the corresponding
Substitute in
Simplify.
Substitute this result into Equation 3 to find
.
How can we check this result?
We can check by finding
and comparing it with the integrand in Equation 1.
First, set up the application of the Product Rule. We'll get 4 terms out of the two products.
Carry out the differentiation.
Simplify.
Does it check?
Yes, differentiation produces the integrand of Equation 1.
The end. If you found this helpful and would recommend that I create more pages like this one, please let me know: Email to John Taylor