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

\frac{d}{d x} (arcsin (2 x^{4}))

\int \frac{x ^{2} + 4 x + 7}{x ^{2} - 6 x + 5} d x

\int_{0}^{t} e^{- x} d x

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

\frac{\partial}{\partial x} (x y ln (x^{2} + y^{2}))

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