# calc

evaluate the integral from 4 to 5.

(x^3 -3x^2 -9)/(x^3 -3x^2)

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1. first off, I'd use partial fractions:

(x^3 -3x^2 -9)/(x^3 -3x^2)
= 1 + 1/x + 3/x^2 - 1/(x-3)

There are no discontinuities on [4,5], so it's straightforward. The integral is just

x + logx - 3/x - log(x-3)

so we get

(5 + log5 - 3/5 - log2)-(4 + log4 - 3/4 - log1)
= 23/20 + log(5/8)

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posted by Steve

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