What is maclaurin x^2ln(9+x) ?

The answer to maclaurin x^2ln(9+x) is 2ln(3)x^2+1/9 x^3-1/162 x^4+1/2187 x^5-1/26244 x^6+\ldots

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maclaurin x^2ln(9+x)

maclaurin

x^{2} ln (9 + x)

2 ln (3) x^{2} + \frac{1}{9} x^{3} - \frac{1}{162} x^{4} + \frac{1}{2187} x^{5} - \frac{1}{26244} x^{6} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{4 ln ( 3 )}{2 !} x^{2} + \frac{\frac{2}{3}}{3 !} x^{3} + \frac{- \frac{4}{27}}{4 !} x^{4} + \frac{\frac{40}{729}}{5 !} x^{5} + \frac{- \frac{20}{729}}{6 !} x^{6} + \dots

Refine

= 2 ln (3) x^{2} + \frac{1}{9} x^{3} - \frac{1}{162} x^{4} + \frac{1}{2187} x^{5} - \frac{1}{26244} x^{6} + \dots

y = 2 x, (32, 8)

\int cos (14 x) cos (5 x) d x

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

\frac{\partial}{\partial y} (y^{5} cos (2 x))

What is maclaurin x^2ln(9+x) ?
The answer to maclaurin x^2ln(9+x) is 2ln(3)x^2+1/9 x^3-1/162 x^4+1/2187 x^5-1/26244 x^6+\ldots