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Chapter 7 Additional Integration Topics

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1 Chapter 7 Additional Integration Topics
Section 3 Integration by Parts

2 Learning Objectives for Section 7.3 Integration by Parts
The student will be able to integrate by parts.

3 An Unsolved Problem In a previous section we said we would return later to the indefinite integral since none of the integration techniques considered up to that time could be used to find an antiderivative for ln x. We now develop a very useful technique, called integration by parts that will enable us to find not only the preceding integral, but also many others.

4 Integration by Parts Integration by parts is based on the product formula for derivatives: We rearrange the above equation to get: Integrating both sides yields which reduces to

5 Integration by Parts Formula
If we let u = f (x) and v = g(x) in the preceding formula, the equation transforms into a more convenient form: Integration by Parts Formula

6 Integration by Parts Formula
This method is useful when the integral on the left side is difficult and changing it into the integral on the right side makes it easier. Remember, we can easily check our results by differentiating our answers to get the original integrals. Let’s look at an example.

7 Example Our previous method of substitution does not work. Examining the left side of the integration by parts formula yields two possibilities. Option 1 u dv Option 2 u dv Let’s try option 1.

8 Example (continued) We have decided to use u = x and dv = e x dx.
This means du = dx and v = e x, yielding This is easy to check by differentiating. Note: Option 2 does not work in this example.

9 Selecting u and dv The product u dv must equal the original integrand.
It must be possible to integrate dv by one of our known methods. The new integral should not be more involved than the original integral For integrals involving x p e ax , try u = x p and dv = e axdx For integrals involving x p (ln x) q, try u = (ln x) q and dv = x p dx

10 Example 2 Find using

11 Example 2 Find using Let u = ln x and dv = x3 dx, yielding du = 1/x dx and v = x4/4 Check by differentiating:

12 Solving the Unsolved Problem
Recall that our original problem was to find We follow the suggestion For integrals involving x p (ln x)q, try u = (ln x)q and dv = x p dx Set p = 0 and q = 1 and choose u = ln x and dv = dx. Then du = 1/x dx and v = x, hence

13 Summary We have developed integration by parts as a useful technique for integrating more complicated functions. Integration by Parts Formula:

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