What is taylor ln(1+3x) ?

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

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taylor ln(1+3x)

taylor

ln (1 + 3 x)

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{3}{1 !} x + \frac{- 9}{2 !} x^{2} + \frac{54}{3 !} x^{3} + \frac{- 486}{4 !} x^{4} + \frac{5832}{5 !} x^{5} + \dots

Refine

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

\frac{d y}{d x} + 2 x y = y + 6 x - 3

\frac{d}{d x} (\frac{2}{3 x})

y^{^{'''}} - 2 y^{^{'}} + y = 0

\int - 9 sin (3 x) d x

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