can anyone help me do this integration (dx)/(x^3 - 2x^2 + 4x - 8) I factored the bottom to (x^2 + 4)(x - 2) and found A = -1/8 B = 1/4 C = 1/8 don't really know what to do next
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Assuming those numbers are correct, the next thing to do is to separate the original fraction into two separate ones, both which can be integrated separately.
hmm the integral of (1/4) / (x^2 + 4) is this (1/4)arctan(x/2) ?
Originally Posted by sleepiiee hmm the integral of (1/4) / (x^2 + 4) is this (1/4)arctan(x/2) ? No. $\displaystyle \frac{\frac{1}{4}}{x^2 + 4} = \frac{1}{8} \cdot \frac{2}{x^2 + 2^2} \, ....$
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