What is taylor (x+1)*ln(x) ?

The answer to taylor (x+1)*ln(x) is 2(x-1)+1/6 (x-1)^3-1/6 (x-1)^4+3/20 (x-1)^5-2/15 (x-1)^6+\ldots

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taylor (x+1)*ln(x)

taylor

(x + 1) \cdot ln (x)

2 (x - 1) + \frac{1}{6} (x - 1)^{3} - \frac{1}{6} (x - 1)^{4} + \frac{3}{20} (x - 1)^{5} - \frac{2}{15} (x - 1)^{6} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{2}{1 !} (x - 1) + \frac{0}{2 !} (x - 1)^{2} + \frac{1}{3 !} (x - 1)^{3} + \frac{- 4}{4 !} (x - 1)^{4} + \frac{18}{5 !} (x - 1)^{5} + \frac{- 96}{6 !} (x - 1)^{6} + \dots

Refine

= 2 (x - 1) + \frac{1}{6} (x - 1)^{3} - \frac{1}{6} (x - 1)^{4} + \frac{3}{20} (x - 1)^{5} - \frac{2}{15} (x - 1)^{6} + \dots

\int \frac{1}{e ^{x}} cos (π) x d x

y^{^{'}} = 10 x y

\int \frac{1}{20} (x + 1)^{2} (x - 3) (x + 4) d x

y^{^{''}} + 4 y^{^{'}} - 4 y = 0

f (x) = \frac{e ^{x}}{x}

What is taylor (x+1)*ln(x) ?
The answer to taylor (x+1)*ln(x) is 2(x-1)+1/6 (x-1)^3-1/6 (x-1)^4+3/20 (x-1)^5-2/15 (x-1)^6+\ldots