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X^3+4x^2-3x+6/(x-1)^2(x^2+3) partial fractions Anybody help?

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  • X^3+4x^2-3x+6/(x-1)^2(x^2+3) partial fractions Anybody help?


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This is what you would input: ... x^2 / ((x + 1) (x^3 + 1)) ... factors on the denominator are linear then the numerators of the partial fractions ... 1 ...
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Positive: 99 %
... x-1)^2}+\frac{C}{(x-1)}+D\right)\,dx$$ I get $$A(x^3 - 3x^2 ... 5/6} = \ln\left((x-1)^{4/3}(x^2 + 2)^{5/6 ... by Partial Fractions $\int\frac{1}{(x+1 ...
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Positive: 96 %

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