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

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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    @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

1 Answers 1

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