maclaurin (x+1)(ln(x+1))

Lösung

maclaurin

(x + 1) (ln (x + 1))

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

Schritte zur Lösung

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Fasse zusammen

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

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