What is maclaurin ln(1+x^3) ?

The answer to maclaurin ln(1+x^3) is x^3-1/2 x^6+1/3 x^9+\ldots

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maclaurin ln(1+x^3)

maclaurin

ln (1 + x^{3})

x^{3} - \frac{1}{2} x^{6} + \frac{1}{3} x^{9} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{0}{2 !} x^{2} + \frac{6}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 360}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{120960}{9 !} x^{9} + \frac{0}{10 !} x^{10} + \dots

Refine

= x^{3} - \frac{1}{2} x^{6} + \frac{1}{3} x^{9} + \dots

\int - 5 csc^{2} (x) + 5 d x

\int (x e^{- x}) d x

\frac{\partial}{\partial y} (x y (x - 2) (y + 3))

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

\frac{\partial}{\partial y} (2 arctan (\frac{x}{y}))

What is maclaurin ln(1+x^3) ?
The answer to maclaurin ln(1+x^3) is x^3-1/2 x^6+1/3 x^9+\ldots