What is taylor ln(1+2x),1 ?

The answer to taylor ln(1+2x),1 is ln(3)+2/3 (x-1)-2/9 (x-1)^2+8/81 (x-1)^3-4/81 (x-1)^4+\ldots

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taylor ln(1+2x),1

taylor

ln (1 + 2 x), 1

ln (3) + \frac{2}{3} (x - 1) - \frac{2}{9} (x - 1)^{2} + \frac{8}{81} (x - 1)^{3} - \frac{4}{81} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= ln (3) + \frac{\frac{2}{3}}{1 !} (x - 1) + \frac{- \frac{4}{9}}{2 !} (x - 1)^{2} + \frac{\frac{16}{27}}{3 !} (x - 1)^{3} + \frac{- \frac{32}{27}}{4 !} (x - 1)^{4} + \dots

Refine

= ln (3) + \frac{2}{3} (x - 1) - \frac{2}{9} (x - 1)^{2} + \frac{8}{81} (x - 1)^{3} - \frac{4}{81} (x - 1)^{4} + \dots

\int cos^{3} (\frac{x}{3}) d x

y = x + 2, y = \frac{1}{x + 1}, x = 0, x = 2

\frac{d y}{d t} = y^{1.01}

\int \frac{ln ( x )}{x ^{1.2}} d x

x \to - 1 lim (2 x + 2)

What is taylor ln(1+2x),1 ?
The answer to taylor ln(1+2x),1 is ln(3)+2/3 (x-1)-2/9 (x-1)^2+8/81 (x-1)^3-4/81 (x-1)^4+\ldots