What is maclaurin 1/(x-3) ?

The answer to maclaurin 1/(x-3) is -1/3-1/9 x-1/27 x^2-1/81 x^3-1/243 x^4+\ldots

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maclaurin 1/(x-3)

maclaurin

\frac{1}{x - 3}

- \frac{1}{3} - \frac{1}{9} x - \frac{1}{27} x^{2} - \frac{1}{81} x^{3} - \frac{1}{243} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= - \frac{1}{3} + \frac{- \frac{1}{9}}{1 !} x + \frac{- \frac{2}{27}}{2 !} x^{2} + \frac{- \frac{2}{27}}{3 !} x^{3} + \frac{- \frac{8}{81}}{4 !} x^{4} + \dots

Refine

= - \frac{1}{3} - \frac{1}{9} x - \frac{1}{27} x^{2} - \frac{1}{81} x^{3} - \frac{1}{243} x^{4} + \dots

\int \frac{2 e ^{2 r}}{7 + e ^{2 r}} d r

\frac{d}{d x} (5^{6^{7^{x}}})

e^{x} y \cdot y^{^{'}} = e^{- y} + e^{- 2 x - y}

f (x) = sin (2 x), - \frac{π}{2}, \frac{π}{6}

\frac{d ^{2}}{d x ^{2}} (11 e^{x^{2}})

What is maclaurin 1/(x-3) ?
The answer to maclaurin 1/(x-3) is -1/3-1/9 x-1/27 x^2-1/81 x^3-1/243 x^4+\ldots