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

The answer to taylor ln^2(1+x) is x^2-x^3+11/12 x^4-5/6 x^5+137/180 x^6+\ldots

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taylor ln^2(1+x)

taylor

ln^{2} (1 + x)

x^{2} - x^{3} + \frac{11}{12} x^{4} - \frac{5}{6} x^{5} + \frac{137}{180} x^{6} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{2}{2 !} x^{2} + \frac{- 6}{3 !} x^{3} + \frac{22}{4 !} x^{4} + \frac{- 100}{5 !} x^{5} + \frac{548}{6 !} x^{6} + \dots

Refine

= x^{2} - x^{3} + \frac{11}{12} x^{4} - \frac{5}{6} x^{5} + \frac{137}{180} x^{6} + \dots

n = 1 \sum \infty \frac{9}{π ^{n}}

\int \frac{1}{361 - x ^{2}} d x

\frac{d y}{d x} - \frac{y}{x} = x e^{x}, y (1) = e - 1

x \to \infty lim (\frac{1}{4 x})

\int_{- 1}^{5} (1 + 3 x) d x

What is taylor ln^2(1+x) ?
The answer to taylor ln^2(1+x) is x^2-x^3+11/12 x^4-5/6 x^5+137/180 x^6+\ldots