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

The answer to taylor e^x+1/(1+x) is 2+3/2 x^2-5/6 x^3+25/24 x^4-119/120 x^5+\ldots

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taylor e^x+1/(1+x)

taylor

e^{x} + \frac{1}{1 + x}

2 + \frac{3}{2} x^{2} - \frac{5}{6} x^{3} + \frac{25}{24} x^{4} - \frac{119}{120} x^{5} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 2 + \frac{0}{1 !} x + \frac{3}{2 !} x^{2} + \frac{- 5}{3 !} x^{3} + \frac{25}{4 !} x^{4} + \frac{- 119}{5 !} x^{5} + \dots

Refine

= 2 + \frac{3}{2} x^{2} - \frac{5}{6} x^{3} + \frac{25}{24} x^{4} - \frac{119}{120} x^{5} + \dots

\frac{d}{d t} (ln (sec (t)))

\frac{d ^{2}}{d x ^{2}} (x^{2} cos (x))

\int \frac{x}{( 3 x ^{2} - 1 )} d x

\int_{1}^{5} \frac{1}{x - 2} d x

n = 1 \sum \infty \frac{n}{1 + 3 n}

What is taylor e^x+1/(1+x) ?
The answer to taylor e^x+1/(1+x) is 2+3/2 x^2-5/6 x^3+25/24 x^4-119/120 x^5+\ldots