What is maclaurin ln(e+x) ?

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

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maclaurin ln(e+x)

maclaurin

ln (e + x)

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

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{\frac{1}{e}}{1 !} x + \frac{- \frac{1}{e ^{2}}}{2 !} x^{2} + \frac{\frac{2}{e ^{3}}}{3 !} x^{3} + \frac{- \frac{6}{e ^{4}}}{4 !} x^{4} + \dots

Refine

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

\int_{0}^{0.5} x (4 - 8 x) d x

\int x cos (n x) d x

1 - 2 x^{2}

x \to 4 lim (x - 5)

\frac{d}{d x} (\frac{x ^{2} + 5}{e ^{x}})

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