What is maclaurin ln(1+e^{-x}) ?

The answer to maclaurin ln(1+e^{-x}) is ln(2)-1/2 x+1/8 x^2-1/192 x^4+1/2880 x^6+\ldots

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maclaurin ln(1+e^{-x})

maclaurin

ln (1 + e^{- x})

ln (2) - \frac{1}{2} x + \frac{1}{8} x^{2} - \frac{1}{192} x^{4} + \frac{1}{2880} x^{6} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= ln (2) + \frac{- \frac{1}{2}}{1 !} x + \frac{\frac{1}{4}}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{- \frac{1}{8}}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{\frac{1}{4}}{6 !} x^{6} + \dots

Refine

= ln (2) - \frac{1}{2} x + \frac{1}{8} x^{2} - \frac{1}{192} x^{4} + \frac{1}{2880} x^{6} + \dots

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

\int sin (t) d t

\frac{d}{d x} (a \cdot x^{2} + b \cdot x + c)

\int \frac{x ^{3} + x + 1}{1 + x ^{2}} d x

(t) e^{4 t}

What is maclaurin ln(1+e^{-x}) ?
The answer to maclaurin ln(1+e^{-x}) is ln(2)-1/2 x+1/8 x^2-1/192 x^4+1/2880 x^6+\ldots