What is taylor 2xln(1+(x^2)/4) ?

The answer to taylor 2xln(1+(x^2)/4) is 1/2 x^3-1/16 x^5+1/96 x^7-1/512 x^9+\ldots

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taylor 2xln(1+(x^2)/4)

taylor

2 x ln (1 + \frac{x ^{2}}{4})

\frac{1}{2} x^{3} - \frac{1}{16} x^{5} + \frac{1}{96} x^{7} - \frac{1}{512} x^{9} + \dots

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( 2 x ln ( 1 + \frac{x ^{2}}{4} ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 2 x ln ( 1 + \frac{x ^{2}}{4} ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 2 x ln ( 1 + \frac{x ^{2}}{4} ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{0}{2 !} x^{2} + \frac{3}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{- \frac{15}{2}}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{\frac{105}{2}}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{- \frac{2835}{4}}{9 !} x^{9} + \frac{0}{10 !} x^{10} + \dots

Refine

= \frac{1}{2} x^{3} - \frac{1}{16} x^{5} + \frac{1}{96} x^{7} - \frac{1}{512} x^{9} + \dots

\int csc^{2} (2 x) cot (2 x) d x

n = 0 \sum \infty \frac{1}{e ^{n}}

\frac{d}{d x} (\frac{x + 1}{x + 4})

\frac{\partial}{\partial x} (tan (x^{3} + y^{2}))

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

What is taylor 2xln(1+(x^2)/4) ?
The answer to taylor 2xln(1+(x^2)/4) is 1/2 x^3-1/16 x^5+1/96 x^7-1/512 x^9+\ldots