What is taylor 3xln(1+x) ?

The answer to taylor 3xln(1+x) is 3x^2-3/2 x^3+x^4-3/4 x^5+3/5 x^6+\ldots

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taylor 3xln(1+x)

taylor

3 x ln (1 + x)

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{6}{2 !} x^{2} + \frac{- 9}{3 !} x^{3} + \frac{24}{4 !} x^{4} + \frac{- 90}{5 !} x^{5} + \frac{432}{6 !} x^{6} + \dots

Refine

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

\int x^{\frac{1}{3}} + (2 x)^{\frac{1}{3}} d x

y^{^{'}} - 2 y = e^{2 t}

y = 13 - x^{2}, y = x^{2} - 5

\frac{d y}{d x} - sin (x) y = 2 sin (x)

\frac{d}{d t} (\frac{d x}{d t}) + \frac{d}{d t} x + 2 x = t^{3}

What is taylor 3xln(1+x) ?
The answer to taylor 3xln(1+x) is 3x^2-3/2 x^3+x^4-3/4 x^5+3/5 x^6+\ldots