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

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

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maclaurin (x^2)/(1-x^2)

maclaurin

\frac{x ^{2}}{1 - x ^{2}}

x^{2} + x^{4} + x^{6} + x^{8} + x^{10} + \dots

Solution steps

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d x} ( \frac{x ^{2}}{1 - x ^{2}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{x ^{2}}{1 - x ^{2}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{x ^{2}}{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{24}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{720}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{40320}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{3628800}{10 !} x^{10} + \dots

Refine

= x^{2} + x^{4} + x^{6} + x^{8} + x^{10} + \dots

\int 15 x^{\frac{1}{2}} d x

\int_{- 2}^{2} x^{2} + 3 x - 4 d x

y^{^{'}} - \frac{6}{x} y = \frac{y ^{4}}{x ^{7}}

\int 4 x^{3} (7 x^{4} + 12)^{3} d x

(6 x^{2} + 4 x)^{2}

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