What is maclaurin ln(1+x^5) ?

The answer to maclaurin ln(1+x^5) is x^5-1/2 x^{10}+\ldots

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maclaurin ln(1+x^5)

maclaurin

ln (1 + x^{5})

x^{5} - \frac{1}{2} x^{10} + \dots

Solution steps

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d x} ( ln ( 1 + x ^{5} ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 1 + x ^{5} ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 1 + x ^{5} ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{0}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{120}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{- 1814400}{10 !} x^{10} + \dots

Refine

= x^{5} - \frac{1}{2} x^{10} + \dots

\int x^{2} 11 + x^{3} d x

f (x) = tan (3 x)

\frac{d}{d x} (- 0.9 x^{2} + 1.7 x + 2.5)

\int x^{3} arcsin (\frac{1}{x}) d x

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

What is maclaurin ln(1+x^5) ?
The answer to maclaurin ln(1+x^5) is x^5-1/2 x^{10}+\ldots