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$\begingroup$

$$\int_{4}^{5}\frac{x^3-3x^2-9}{x^3-3x^2}$$

I used long division to get: $$\int_{4}^{5}1+\frac{9}{x^3-3x^2}$$

Factored the denominator:

$$x^2(x-3)$$

Broke the rational function into partial fractions:

$$\frac{9}{x^2(x-3)} = \frac{A}{x-3} + \frac{B}{x} + \frac{C}{x^2}$$

Solved for A, B, and C and got: $$A = 1,\quad B = -1,\quad C = -3$$

Substituted the values into the equation:

$$\int_{4}^{5}1 + \frac{1}{x-3} - \frac{1}{x} - \frac{3}{x^2}$$

Found the anti-derivative: $$[x + \ln|x-3| - \ln|x| - 3\ln|x^2|]_{4}^{5}$$

I keep getting the answer wrong so at this point, I don't know what I'm doing wrong. Any help would be appreciated.

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    The integral of $\frac{1}{x^2}$ is not $\ln|x^2|$. $$\int\frac{1}{x^2}\,dx = \int x^{-2}\,dx = -x^{-1}+C = -\frac{1}{x}+C.$$2012-03-05
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    So it should look something like this: $[x + \ln|x-3| - \ln|x| + 3/x]_{4}^{5}$2012-03-05
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    Exactly; then you should get the right answer.2012-03-05
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    I solved the equation and it gave me the answer of 1.32 while wolfram puts it at 0.68. What gives?2012-03-05
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    You have a sign error in your first step that I did not notice: you should have $$\frac{x^3-3x^2-9}{x^3-3x^2} = 1 - \frac{9}{x^3-3x^2};$$ instead, you have $$\frac{x^3-3x^2-9}{x^3-3x^2} = 1 + \frac{9}{x^3-3x^2}.$$2012-03-05
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    @mathisnotmyforte: It is an important habit to write the$\textrm{d}x$ for each integral. Leaving off the $\textrm{d}x$ can lead to easy mistakes.2012-03-05

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