What is maclaurin ln(x-1) ?

The answer to maclaurin ln(x-1) is Does not exist-x-1/2 x^2-1/3 x^3-1/4 x^4+\ldots

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maclaurin ln(x-1)

maclaurin

ln (x - 1)

Does not exist - x - \frac{1}{2} x^{2} - \frac{1}{3} x^{3} - \frac{1}{4} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= Does not exist + \frac{\frac{d}{d x} ( ln ( x - 1 ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( x - 1 ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( x - 1 ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= Does not exist + \frac{- 1}{1 !} x + \frac{- 1}{2 !} x^{2} + \frac{- 2}{3 !} x^{3} + \frac{- 6}{4 !} x^{4} + \dots

Refine

= Does not exist - x - \frac{1}{2} x^{2} - \frac{1}{3} x^{3} - \frac{1}{4} x^{4} + \dots

\frac{d y}{d t} = \frac{e ^{y} + 1}{e ^{y}}

x \to 2 - lim (2 x - 3)

y = 1 - cot (2 x)

x \frac{d y}{d x} = \frac{1 - y ^{2}}{2 y}

\int_{0}^{4} \frac{4 t}{( t - 5 ) ^{2}} d t

What is maclaurin ln(x-1) ?
The answer to maclaurin ln(x-1) is Does not exist-x-1/2 x^2-1/3 x^3-1/4 x^4+\ldots