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

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

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

taylor

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

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

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

\frac{\partial}{\partial u} (u - uv)

\frac{2}{x ^{4}}

f (y) = y ln (1 + e^{y})

\int \frac{cos ( x )}{sin ^{3} ( x ) - cos ^{3} ( x )} d x

\frac{d}{d x} (sin (2 x) + 2 x cos (2 x))

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