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

The answer to taylor 1/((1+x)^2),1 is 1/4-1/4 (x-1)+3/16 (x-1)^2-1/8 (x-1)^3+5/64 (x-1)^4+\ldots

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taylor 1/((1+x)^2),1

taylor

\frac{1}{( 1 + x ) ^{2}}, 1

\frac{1}{4} - \frac{1}{4} (x - 1) + \frac{3}{16} (x - 1)^{2} - \frac{1}{8} (x - 1)^{3} + \frac{5}{64} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

= \frac{1}{4} + \frac{\frac{d}{d x} ( \frac{1}{( 1 + x ) ^{2}} ) ( 1 )}{1 !} (x - 1) + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{( 1 + x ) ^{2}} ) ( 1 )}{2 !} (x - 1)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{( 1 + x ) ^{2}} ) ( 1 )}{3 !} (x - 1)^{3} + \dots

Evaluate Derivatives

= \frac{1}{4} + \frac{- \frac{1}{4}}{1 !} (x - 1) + \frac{\frac{3}{8}}{2 !} (x - 1)^{2} + \frac{- \frac{3}{4}}{3 !} (x - 1)^{3} + \frac{\frac{15}{8}}{4 !} (x - 1)^{4} + \dots

Refine

= \frac{1}{4} - \frac{1}{4} (x - 1) + \frac{3}{16} (x - 1)^{2} - \frac{1}{8} (x - 1)^{3} + \frac{5}{64} (x - 1)^{4} + \dots

\frac{d}{d x} (5 x)

\int 4 x^{2} + 1 d x

5 x^{2} + 2 x + 3

\frac{\partial}{\partial x} (x^{2} - 4 x y + y^{2} + 6 y + 3)

\int tan^{5} (x) sec^{11} (x) d x

What is taylor 1/((1+x)^2),1 ?
The answer to taylor 1/((1+x)^2),1 is 1/4-1/4 (x-1)+3/16 (x-1)^2-1/8 (x-1)^3+5/64 (x-1)^4+\ldots