What is maclaurin ln(1/(1-x)) ?

The answer to maclaurin ln(1/(1-x)) is x+1/2 x^2+1/3 x^3+1/4 x^4+1/5 x^5+\ldots

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maclaurin ln(1/(1-x))

maclaurin

ln (\frac{1}{1 - x})

x + \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \frac{1}{4} x^{4} + \frac{1}{5} x^{5} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{1}{2 !} x^{2} + \frac{2}{3 !} x^{3} + \frac{6}{4 !} x^{4} + \frac{24}{5 !} x^{5} + \dots

Refine

= x + \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \frac{1}{4} x^{4} + \frac{1}{5} x^{5} + \dots

\frac{d}{d x} (cos (10 x) - 0.1 sin (10 x))

\int \frac{2}{x ^{2} + 3} d x

(x^{2} y^{3} - \frac{1}{1 + 9 x ^{2}}) d x + (x^{3} y^{2}) d y = 0

\frac{d}{d x} (sin^{2} (x) - cos^{2} (x))

ln (2 x + 1)

What is maclaurin ln(1/(1-x)) ?
The answer to maclaurin ln(1/(1-x)) is x+1/2 x^2+1/3 x^3+1/4 x^4+1/5 x^5+\ldots