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Understanding Integration by Parts: A Key Technique for Advanced Integration Problems

In this section, we introduce the concept of integration by parts, a crucial technique for evaluating complex integrals. Initially, we address the integration of functions like ln(x), which cannot be solved using previous methods. The integration by parts formula, derived from the product rule for derivatives, allows us to transform difficult integrals into simpler ones. Through step-by-step examples, we demonstrate how to select components of the integral (u and dv) and ensure the new integral is manageable. Mastering this technique significantly broadens the scope of integrable functions.

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Understanding Integration by Parts: A Key Technique for Advanced Integration Problems

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  1. Chapter 7Additional 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. Option 1 Option 2 u dv u dv Example Our previous method of substitution does not work. Examining the left side of the integration by parts formula yields two possibilities. 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 pe 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/xdx 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/xdx 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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