What is maclaurin 1/(x+5) ?

The answer to maclaurin 1/(x+5) is 1/5-1/25 x+1/125 x^2-1/625 x^3+1/3125 x^4+\ldots

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maclaurin 1/(x+5)

maclaurin

\frac{1}{x + 5}

\frac{1}{5} - \frac{1}{25} x + \frac{1}{125} x^{2} - \frac{1}{625} x^{3} + \frac{1}{3125} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= \frac{1}{5} + \frac{\frac{d}{d x} ( \frac{1}{x + 5} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{x + 5} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{x + 5} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= \frac{1}{5} + \frac{- \frac{1}{25}}{1 !} x + \frac{\frac{2}{125}}{2 !} x^{2} + \frac{- \frac{6}{625}}{3 !} x^{3} + \frac{\frac{24}{3125}}{4 !} x^{4} + \dots

Refine

= \frac{1}{5} - \frac{1}{25} x + \frac{1}{125} x^{2} - \frac{1}{625} x^{3} + \frac{1}{3125} x^{4} + \dots

\frac{d}{d x} (x + i y (x))

(x + y)^{2} d x + (2 x y + x^{2} - 5) d y = 0, y (1) = 1

\int_{0}^{8} \frac{1}{x ^{2} + 16} d x

f (X) = \frac{ln ( 2 )}{ln ( X )}

- \frac{d}{d x} (\frac{d y}{d x}) + 400 y = 0

What is maclaurin 1/(x+5) ?
The answer to maclaurin 1/(x+5) is 1/5-1/25 x+1/125 x^2-1/625 x^3+1/3125 x^4+\ldots