What is maclaurin ln(1+x^2) ?

The answer to maclaurin ln(1+x^2) is x^2-1/2 x^4+1/3 x^6-1/4 x^8+1/5 x^{10}+\ldots

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maclaurin ln(1+x^2)

maclaurin

ln (1 + x^{2})

x^{2} - \frac{1}{2} x^{4} + \frac{1}{3} x^{6} - \frac{1}{4} x^{8} + \frac{1}{5} x^{10} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{2}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{- 12}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{240}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{- 10080}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{725760}{10 !} x^{10} + \dots

Refine

= x^{2} - \frac{1}{2} x^{4} + \frac{1}{3} x^{6} - \frac{1}{4} x^{8} + \frac{1}{5} x^{10} + \dots

f (x) = e^{2} + ln (7)

(\frac{1 - cos ( x )}{x})^{^{'}}

\int \frac{x - 2}{x ^{2} + x + 1} d x

\frac{d}{d x} (- csc^{2} (x) - cos (2 x) \cdot 2)

(y^{2} + 1) d x = y (sec^{2} (x)) d y

What is maclaurin ln(1+x^2) ?
The answer to maclaurin ln(1+x^2) is x^2-1/2 x^4+1/3 x^6-1/4 x^8+1/5 x^{10}+\ldots