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

Factored the denominator:


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 $\ln|x^2|$. $$\int\frac{1}{x^2}\,dx = \int x^{-2}\,dx = -x^{-1}+C = -\frac{1}{x}+C.$$ – Arturo Magidin Mar 5 '12 at 1:53
So it should look something like this: $[x + \ln|x-3| - \ln|x| + 3/x]_{4}^{5}$ – mathisnotmyforte Mar 5 '12 at 2:07
Exactly; then you should get the right answer. – Arturo Magidin Mar 5 '12 at 2:13
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 $$\frac{x^3-3x^2-9}{x^3-3x^2} = 1 + \frac{9}{x^3-3x^2}.$$ – Arturo Magidin Mar 5 '12 at 2:22
@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. – JavaMan Mar 5 '12 at 3:56

Your only errors are a trivial one, $$\frac{x^3-3x^2 -9}{x^3-3x^2} = 1 - \frac{9}{x^3-3x^2}$$ (you have a $+$ instead), and a somewhat more serious one that lies in evaluating the integral of $\frac{3}{x^2}$. The latter is not $3\ln(x^2)$. First, remember that the integral $$\int\frac{dx}{f(x)}$$ is almost never equal to $\ln|f(x)|$. And second, remember that $$\int x^n\,dx = \left\{\begin{array}{ll} \frac{1}{n+1}x^{n+1} +C &\text{if }n\neq -1;\\ \strut\\ \ln|x| + C &\text{if }n=-1. \end{array}\right.$$

So $$\begin{align*} \int_4^5 -\frac{3}{x^2}\,dx &= -3\int_4^5 x^{-2}\,dx\\ &= -3\left(\left. \frac{1}{-2+1}x^{-2+1}\right|_{4}^5\right)\\ &= -3\left(\left.\frac{1}{-1}x^-1\right|_{4}^5\right)\\ &= -3\left(\left. -\frac{1}{x}\right|_{4}^5\right)\\ &= -3\left( -\frac{1}{5} + \frac{1}{4}\right). \end{align*}$$

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