What is taylor (x-1)ln(x),1 ?

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

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taylor (x-1)ln(x),1

taylor

(x - 1) ln (x), 1

(x - 1)^{2} - \frac{1}{2} (x - 1)^{3} + \frac{1}{3} (x - 1)^{4} - \frac{1}{4} (x - 1)^{5} + \frac{1}{5} (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{0}{1 !} (x - 1) + \frac{2}{2 !} (x - 1)^{2} + \frac{- 3}{3 !} (x - 1)^{3} + \frac{8}{4 !} (x - 1)^{4} + \frac{- 30}{5 !} (x - 1)^{5} + \frac{144}{6 !} (x - 1)^{6} + \dots

Refine

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

\frac{\partial}{\partial x} (sin (x))

\int x^{2} + \frac{3}{4} d x

\int y sin (y^{2}) d y

\int \frac{24 x}{3 3 x ^{2} - 4} d x

\int 5 sin^{2} (x co) s^{2} x d x

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