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Problem 18 Easy Difficulty

Use the Table of Integrals on Reference Pages 6-10 to evaluate the integral.

$ \displaystyle \int \frac{dx}{2x^3 - 3x^2} $

Answer

$\frac{1}{3 x}+\frac{2}{9} \ln \left|\frac{2 x-3}{x}\right|+C$

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Video Transcript

Okay, This question wants us to compute this integral. So to do this, we need to find a foreman or table that looks like this. But that's not gonna work out unless if we do some pre processing here. So well, it's factor out our common factor of X squared on the bottom here. And no, if we look at our table in the back of the book, we see this formula, which is exactly the form that are integral is in. We just have to flip some signs. So we get integral of D X over X squared times, negative three plus two X and now that's exactly in this form. So we just gotta value at this formula with our values of A and B. So So a is the constant term, which is negative three and B equals our term in front of the X, which is to So now we just need to plug into our formula which says that it's negative one divided by negative three times are variable x plus. See over a squared times the natural log of a plus b, u or B X. In this case divided by X plus c and then we can just do a little bit of simplification and get won over three X plus 2/9 times the natural log of two X minus three over X plus C, which is our final answer.

University of Michigan - Ann Arbor
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