What is maclaurin 1/(x+2) ?

The answer to maclaurin 1/(x+2) is 1/2-1/4 x+1/8 x^2-1/16 x^3+1/32 x^4+\ldots

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maclaurin 1/(x+2)

maclaurin

\frac{1}{x + 2}

\frac{1}{2} - \frac{1}{4} x + \frac{1}{8} x^{2} - \frac{1}{16} x^{3} + \frac{1}{32} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

= \frac{1}{2} - \frac{1}{4} x + \frac{1}{8} x^{2} - \frac{1}{16} x^{3} + \frac{1}{32} x^{4} + \dots

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

\int x^{8} e^{x} d x

(x + 2) \frac{d y}{d x} + y = ln (x), y (1) = 20

x, x^{2}, 0, 1

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

What is maclaurin 1/(x+2) ?
The answer to maclaurin 1/(x+2) is 1/2-1/4 x+1/8 x^2-1/16 x^3+1/32 x^4+\ldots