What is maclaurin (x+1)(ln(x+1)) ?

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

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

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

Solution steps

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

Refine

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

e^{2 x}, e^{5 x}, x = 1

\frac{\partial}{\partial x} (x 1 + y^{9})

\frac{\partial}{\partial x} (5 e^{- x} cos (y))

\frac{d}{d x} (\frac{1}{1 + 4 x ^{2} + 9 x ^{4}})

\frac{d}{d θ} (2 - 2 sin (θ))

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