What is taylor 1/(x^2),-1 ?

The answer to taylor 1/(x^2),-1 is 1+2(x+1)+3(x+1)^2+4(x+1)^3+5(x+1)^4+\ldots

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taylor 1/(x^2),-1

taylor

\frac{1}{x ^{2}}, - 1

1 + 2 (x + 1) + 3 (x + 1)^{2} + 4 (x + 1)^{3} + 5 (x + 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{2}{1 !} (x + 1) + \frac{6}{2 !} (x + 1)^{2} + \frac{24}{3 !} (x + 1)^{3} + \frac{120}{4 !} (x + 1)^{4} + \dots

Refine

= 1 + 2 (x + 1) + 3 (x + 1)^{2} + 4 (x + 1)^{3} + 5 (x + 1)^{4} + \dots

\int (\frac{6 - 2 x}{8 - 4 x - 4 x ^{2}}) d x

\frac{d}{d x} (\frac{7}{x ^{4} + 3})

\frac{\partial}{\partial x} ((x - 1)^{2} + 8 (y - 1)^{2})

x \to \frac{π}{4} lim ((tan (x))^{t a n (2 x)})

x \to 5 + lim (ln (x^{2} - 25))

What is taylor 1/(x^2),-1 ?
The answer to taylor 1/(x^2),-1 is 1+2(x+1)+3(x+1)^2+4(x+1)^3+5(x+1)^4+\ldots