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

The answer to taylor 1/(x(x+1)) is 1/2-3/4 (x-1)+7/8 (x-1)^2-15/16 (x-1)^3+31/32 (x-1)^4+\ldots

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taylor 1/(x(x+1))

taylor

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

\frac{1}{2} - \frac{3}{4} (x - 1) + \frac{7}{8} (x - 1)^{2} - \frac{15}{16} (x - 1)^{3} + \frac{31}{32} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{2} + \frac{- \frac{3}{4}}{1 !} (x - 1) + \frac{\frac{7}{4}}{2 !} (x - 1)^{2} + \frac{- \frac{45}{8}}{3 !} (x - 1)^{3} + \frac{\frac{93}{4}}{4 !} (x - 1)^{4} + \dots

Refine

= \frac{1}{2} - \frac{3}{4} (x - 1) + \frac{7}{8} (x - 1)^{2} - \frac{15}{16} (x - 1)^{3} + \frac{31}{32} (x - 1)^{4} + \dots

\int \frac{3 x ^{2} + 4 x + 1}{3 x ^{3} + 6 x ^{2} + 3 x + 5} d x

y = x^{2} sin^{6} (x) + x cos^{- 4} (x)

\frac{d}{d x} (\frac{x - 1}{x ^{2} + 1})

\frac{d}{d x} (\frac{3 x ^{4}}{4})

\int \frac{2 x - 1}{x ^{2} - x - 12} d x

What is taylor 1/(x(x+1)) ?
The answer to taylor 1/(x(x+1)) is 1/2-3/4 (x-1)+7/8 (x-1)^2-15/16 (x-1)^3+31/32 (x-1)^4+\ldots