What is taylor ln(1+9x) ?

The answer to taylor ln(1+9x) is 9x-81/2 x^2+243x^3-6561/4 x^4+59049/5 x^5+\ldots

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taylor ln(1+9x)

taylor

ln (1 + 9 x)

9 x - \frac{81}{2} x^{2} + 243 x^{3} - \frac{6561}{4} x^{4} + \frac{59049}{5} x^{5} + \dots

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( ln ( 1 + 9 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 1 + 9 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 1 + 9 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{9}{1 !} x + \frac{- 81}{2 !} x^{2} + \frac{1458}{3 !} x^{3} + \frac{- 39366}{4 !} x^{4} + \frac{1417176}{5 !} x^{5} + \dots

Refine

= 9 x - \frac{81}{2} x^{2} + 243 x^{3} - \frac{6561}{4} x^{4} + \frac{59049}{5} x^{5} + \dots

\frac{\partial}{\partial a} (e^{a} sin (a + b (a)))

\int \frac{x ^{5}}{x ^{2} + 2} d x

ln (x)

sin (x) 2

\int e^{5 x} sin (2 x) d x

What is taylor ln(1+9x) ?
The answer to taylor ln(1+9x) is 9x-81/2 x^2+243x^3-6561/4 x^4+59049/5 x^5+\ldots