What is taylor 1/(2x+5) ?

The answer to taylor 1/(2x+5) is 1/5-2/25 x+4/125 x^2-8/625 x^3+16/3125 x^4+\ldots

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taylor 1/(2x+5)

taylor

\frac{1}{2 x + 5}

\frac{1}{5} - \frac{2}{25} x + \frac{4}{125} x^{2} - \frac{8}{625} x^{3} + \frac{16}{3125} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{5} + \frac{- \frac{2}{25}}{1 !} x + \frac{\frac{8}{125}}{2 !} x^{2} + \frac{- \frac{48}{625}}{3 !} x^{3} + \frac{\frac{384}{3125}}{4 !} x^{4} + \dots

Refine

= \frac{1}{5} - \frac{2}{25} x + \frac{4}{125} x^{2} - \frac{8}{625} x^{3} + \frac{16}{3125} x^{4} + \dots

2 x^{2} - 3 x + 5

f (x) = \frac{sin ( 5 x )}{3 x}

\int 16 x^{5} d x

\int (8 x (x^{7} - \frac{1}{8})) d x

\int y e^{- x} d x

What is taylor 1/(2x+5) ?
The answer to taylor 1/(2x+5) is 1/5-2/25 x+4/125 x^2-8/625 x^3+16/3125 x^4+\ldots