What is maclaurin (3x^2-4x+1)/(x^3+1) ?

The answer to maclaurin (3x^2-4x+1)/(x^3+1) is 1-4x+3x^2-x^3+4x^4+\ldots

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maclaurin (3x^2-4x+1)/(x^3+1)

maclaurin

\frac{3 x ^{2} - 4 x + 1}{x ^{3} + 1}

1 - 4 x + 3 x^{2} - x^{3} + 4 x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= 1 + \frac{\frac{d}{d x} ( \frac{3 x ^{2} - 4 x + 1}{x ^{3} + 1} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{3 x ^{2} - 4 x + 1}{x ^{3} + 1} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{3 x ^{2} - 4 x + 1}{x ^{3} + 1} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{- 4}{1 !} x + \frac{6}{2 !} x^{2} + \frac{- 6}{3 !} x^{3} + \frac{96}{4 !} x^{4} + \dots

Refine

= 1 - 4 x + 3 x^{2} - x^{3} + 4 x^{4} + \dots

\int \frac{1}{x ^{3} x ^{2} - 2} d x

f (x) = x^{3} - 2 x^{2}, g (x) = x^{2} - 2 x

\frac{\partial}{\partial x} (2 x^{2} + 1)

x \to 6 lim (\frac{5}{6} + 9 (- \frac{2}{3}))

\frac{d}{d x} (ln (x + 7 + x^{2}))

What is maclaurin (3x^2-4x+1)/(x^3+1) ?
The answer to maclaurin (3x^2-4x+1)/(x^3+1) is 1-4x+3x^2-x^3+4x^4+\ldots