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Section 6 .1, problem 19.
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We're dealing with a problem where we need to do integration by part.
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So you've got calculus and by very nature, you've got calculus algebra, the fact that it's a definite integral, you're going to have arithmetic too in this problem.
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So in this case, we normally let you be the polynomial.
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The exception is typically when you have a logarithm term, you're going to let you be that logarithm term because when you differentiate the logarithm function, the log goes away.
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And you have something that is easier to work with, a rational function.
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Then we can let dv equal r -cubed dr.
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So when we integrate that, by integration we'll get v is equal to r -to -the -fourth over four.
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So that tells us that let's work on the indefinite interval first.
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So to integrate r -cube, natural log of r -dr is going to be uv...