What is maclaurin ln(1-x^2) ?

The answer to maclaurin ln(1-x^2) is -x^2-1/2 x^4-1/3 x^6-1/4 x^8-1/5 x^{10}+\ldots

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maclaurin ln(1-x^2)

maclaurin

ln (1 - x^{2})

- x^{2} - \frac{1}{2} x^{4} - \frac{1}{3} x^{6} - \frac{1}{4} x^{8} - \frac{1}{5} x^{10} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{- 2}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{- 12}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 240}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{- 10080}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{- 725760}{10 !} x^{10} + \dots

Refine

= - x^{2} - \frac{1}{2} x^{4} - \frac{1}{3} x^{6} - \frac{1}{4} x^{8} - \frac{1}{5} x^{10} + \dots

\int t^{2} e^{\frac{t}{3}} d t

x \to 4 lim (\frac{5}{x - 4})

x \to \infty lim (\frac{( e ^{- x y - x + y} )}{x ^{2} y ^{2} + x y})

y = x^{2} - 3, y = x + 3

\int \frac{4 x - 11}{x ^{3} - 9 x ^{2}} d x

What is maclaurin ln(1-x^2) ?
The answer to maclaurin ln(1-x^2) is -x^2-1/2 x^4-1/3 x^6-1/4 x^8-1/5 x^{10}+\ldots