What is maclaurin ln(e+x^2) ?

The answer to maclaurin ln(e+x^2) is 1+1/e x^2-1/(2e^2)x^4+1/(3e^3)x^6-1/(4e^4)x^8+\ldots

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maclaurin ln(e+x^2)

maclaurin

ln (e + x^{2})

1 + \frac{1}{e} x^{2} - \frac{1}{2 e ^{2}} x^{4} + \frac{1}{3 e ^{3}} x^{6} - \frac{1}{4 e ^{4}} x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{\frac{2}{e}}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{- \frac{12}{e ^{2}}}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{\frac{240}{e ^{3}}}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{- \frac{10080}{e ^{4}}}{8 !} x^{8} + \dots

Refine

= 1 + \frac{1}{e} x^{2} - \frac{1}{2 e ^{2}} x^{4} + \frac{1}{3 e ^{3}} x^{6} - \frac{1}{4 e ^{4}} x^{8} + \dots

\int cos^{2} (x) x d x

(\frac{1}{x}), 2

\frac{1}{s ^{2} ( s ^{2} + 4 )}

y = 3 x - 6, (2, 4)

\frac{d}{d x} (\frac{2 x ^{3} - 5 x ^{2} + 4 x}{( x - 1 ) ^{2}})

What is maclaurin ln(e+x^2) ?
The answer to maclaurin ln(e+x^2) is 1+1/e x^2-1/(2e^2)x^4+1/(3e^3)x^6-1/(4e^4)x^8+\ldots