What is maclaurin ln(1+2x) ?

The answer to maclaurin ln(1+2x) is 2x-2x^2+8/3 x^3-4x^4+32/5 x^5+\ldots

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maclaurin ln(1+2x)

maclaurin

ln (1 + 2 x)

2 x - 2 x^{2} + \frac{8}{3} x^{3} - 4 x^{4} + \frac{32}{5} x^{5} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{2}{1 !} x + \frac{- 4}{2 !} x^{2} + \frac{16}{3 !} x^{3} + \frac{- 96}{4 !} x^{4} + \frac{768}{5 !} x^{5} + \dots

Refine

= 2 x - 2 x^{2} + \frac{8}{3} x^{3} - 4 x^{4} + \frac{32}{5} x^{5} + \dots

\int ln^{3} (2 x) d x

\frac{1}{x - 3}

\frac{d}{d x} (ln (y x))

x y^{^{'}} - (1 + x) y = x y^{2}

x \to 1 lim (\frac{x - 1}{x + 24 - 5})

What is maclaurin ln(1+2x) ?
The answer to maclaurin ln(1+2x) is 2x-2x^2+8/3 x^3-4x^4+32/5 x^5+\ldots