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Partial fraction decomposition (2x^2-x+4)/(x^3+4x) I got a=1 b=1 c=-1 which is wrong I'm sure
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its correct i think
\(\large\tt \color{black}{\dfrac{2x^2-x+4}{x^3+4x}=\dfrac{1}{x}+\dfrac{x-1}{x^2+4}}\)
Ok so I'm glad I got that portion of the problem right lol thanks
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