What is maclaurin ln(4+x^2) ?

The answer to maclaurin ln(4+x^2) is 2ln(2)+1/4 x^2-1/32 x^4+1/192 x^6-1/1024 x^8+\ldots

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maclaurin ln(4+x^2)

maclaurin

ln (4 + x^{2})

2 ln (2) + \frac{1}{4} x^{2} - \frac{1}{32} x^{4} + \frac{1}{192} x^{6} - \frac{1}{1024} x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 2 ln (2) + \frac{0}{1 !} x + \frac{\frac{1}{2}}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{- \frac{3}{4}}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{\frac{15}{4}}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{- \frac{315}{8}}{8 !} x^{8} + \dots

Refine

= 2 ln (2) + \frac{1}{4} x^{2} - \frac{1}{32} x^{4} + \frac{1}{192} x^{6} - \frac{1}{1024} x^{8} + \dots

\int_{1}^{2} 15 4 x^{2} - 3 d x

f (x) = x^{2} (1 - 6 x)

n = 0 \sum \infty 7 (- \frac{1}{7})^{n}

\frac{\partial}{\partial x} (- x sin (y) + e^{x})

y^{^{''}} - 4 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 7

What is maclaurin ln(4+x^2) ?
The answer to maclaurin ln(4+x^2) is 2ln(2)+1/4 x^2-1/32 x^4+1/192 x^6-1/1024 x^8+\ldots