What is maclaurin ln(4+x) ?

The answer to maclaurin ln(4+x) is 2ln(2)+1/4 x-1/32 x^2+1/192 x^3-1/1024 x^4+\ldots

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maclaurin ln(4+x)

maclaurin

ln (4 + x)

2 ln (2) + \frac{1}{4} x - \frac{1}{32} x^{2} + \frac{1}{192} x^{3} - \frac{1}{1024} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= 2 ln (2) + \frac{\frac{d}{d x} ( ln ( 4 + x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 4 + x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 4 + x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 2 ln (2) + \frac{\frac{1}{4}}{1 !} x + \frac{- \frac{1}{16}}{2 !} x^{2} + \frac{\frac{1}{32}}{3 !} x^{3} + \frac{- \frac{3}{128}}{4 !} x^{4} + \dots

Refine

= 2 ln (2) + \frac{1}{4} x - \frac{1}{32} x^{2} + \frac{1}{192} x^{3} - \frac{1}{1024} x^{4} + \dots

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

\frac{d}{d x} (e^{2 s i n (x)})

5 x^{2} y^{^{'}} = y^{^{'}} + 9 x e^{- y}

\int ln (x) x^{7} d x

y = - 2 x^{4} + 7 x e^{x}

What is maclaurin ln(4+x) ?
The answer to maclaurin ln(4+x) is 2ln(2)+1/4 x-1/32 x^2+1/192 x^3-1/1024 x^4+\ldots