What is taylor 1/(1+e^{x^2)} ?

The answer to taylor 1/(1+e^{x^2)} is 1/2-1/4 x^2+1/48 x^6+\ldots

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taylor 1/(1+e^{x^2)}

taylor

\frac{1}{1 + e ^{x^{2}}}

\frac{1}{2} - \frac{1}{4} x^{2} + \frac{1}{48} x^{6} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{2} + \frac{0}{1 !} x + \frac{- \frac{1}{2}}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{15}{6 !} x^{6} + \dots

Refine

= \frac{1}{2} - \frac{1}{4} x^{2} + \frac{1}{48} x^{6} + \dots

\int \frac{1}{25 - 36 x ^{2}} d x

\int 13 tan^{2} (x) sec^{3} (x) d x

\int - 2 x^{2} + 8 x d x

y^{^{''}} - 3 y^{^{'}} + y = 0

y = 3 x^{3} - 3 x

What is taylor 1/(1+e^{x^2)} ?
The answer to taylor 1/(1+e^{x^2)} is 1/2-1/4 x^2+1/48 x^6+\ldots