What is taylor ln(((1+x))/(1-x)) ?

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

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taylor ln(((1+x))/(1-x))

taylor

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

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{2}{1 !} x + \frac{0}{2 !} x^{2} + \frac{4}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{48}{5 !} x^{5} + \dots

Refine

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

\frac{d}{d x} (- (x - 3)^{2} (1 - x)^{2})

\frac{\partial}{\partial x} ((4 x + 4) (ln (8 x y)))

\frac{d}{d x} (A x^{2} e^{- 2 x})

\int (sec^{2} (x) - 5) d x

f (x) = (1 + 5 x)^{9}, at x = 0

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