What is maclaurin ln(1+5x) ?

The answer to maclaurin ln(1+5x) is 5x-25/2 x^2+125/3 x^3-625/4 x^4+625x^5+\ldots

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maclaurin ln(1+5x)

maclaurin

ln (1 + 5 x)

5 x - \frac{25}{2} x^{2} + \frac{125}{3} x^{3} - \frac{625}{4} x^{4} + 625 x^{5} + \dots

Solution steps

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d x} ( ln ( 1 + 5 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 1 + 5 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 1 + 5 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{5}{1 !} x + \frac{- 25}{2 !} x^{2} + \frac{250}{3 !} x^{3} + \frac{- 3750}{4 !} x^{4} + \frac{75000}{5 !} x^{5} + \dots

Refine

= 5 x - \frac{25}{2} x^{2} + \frac{125}{3} x^{3} - \frac{625}{4} x^{4} + 625 x^{5} + \dots

\int_{0}^{1} x 1 - x^{4} d x

\int x x^{\frac{3}{2}} - 1 d x

\frac{d}{d x} (lo g_{5} (2 x^{2} + 1))

(2 x + \frac{2}{y}) + (\frac{( 2 x ^{2} )}{y} + \frac{2 x}{y ^{2}}) y^{^{'}} = 0

f (x) = x^{4} - 10 x^{2} + 9

What is maclaurin ln(1+5x) ?
The answer to maclaurin ln(1+5x) is 5x-25/2 x^2+125/3 x^3-625/4 x^4+625x^5+\ldots