What is taylor xln(x),1 ?

The answer to taylor xln(x),1 is (x-1)+1/2 (x-1)^2-1/6 (x-1)^3+1/12 (x-1)^4-1/20 (x-1)^5+\ldots

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taylor xln(x),1

taylor

x ln (x), 1

(x - 1) + \frac{1}{2} (x - 1)^{2} - \frac{1}{6} (x - 1)^{3} + \frac{1}{12} (x - 1)^{4} - \frac{1}{20} (x - 1)^{5} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

= (x - 1) + \frac{1}{2} (x - 1)^{2} - \frac{1}{6} (x - 1)^{3} + \frac{1}{12} (x - 1)^{4} - \frac{1}{20} (x - 1)^{5} + \dots

x \to 0 lim (arctan (x^{2} + 1))

\int_{0}^{1} \int_{0}^{2} x y e^{x y^{2}} d x d y

\int (e^{2 x} + 4)^{- 1} d x

\int_{l n (4)}^{l n (25)} e^{\frac{x}{2}} d x

x \to 0 lim (- \frac{5}{1 - 5 x})

What is taylor xln(x),1 ?
The answer to taylor xln(x),1 is (x-1)+1/2 (x-1)^2-1/6 (x-1)^3+1/12 (x-1)^4-1/20 (x-1)^5+\ldots