What is maclaurin 1/(5x+7) ?

The answer to maclaurin 1/(5x+7) is 1/7-5/49 x+25/343 x^2-125/2401 x^3+625/16807 x^4+\ldots

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maclaurin 1/(5x+7)

maclaurin

\frac{1}{5 x + 7}

\frac{1}{7} - \frac{5}{49} x + \frac{25}{343} x^{2} - \frac{125}{2401} x^{3} + \frac{625}{16807} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= \frac{1}{7} + \frac{\frac{d}{d x} ( \frac{1}{5 x + 7} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{5 x + 7} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{5 x + 7} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= \frac{1}{7} + \frac{- \frac{5}{49}}{1 !} x + \frac{\frac{50}{343}}{2 !} x^{2} + \frac{- \frac{750}{2401}}{3 !} x^{3} + \frac{\frac{15000}{16807}}{4 !} x^{4} + \dots

Refine

= \frac{1}{7} - \frac{5}{49} x + \frac{25}{343} x^{2} - \frac{125}{2401} x^{3} + \frac{625}{16807} x^{4} + \dots

y^{^{'}} = 2 y - y^{3}

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

y^{^{'}} = x + 3 y

(1 - 2 x)^{\frac{1}{2}}

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

What is maclaurin 1/(5x+7) ?
The answer to maclaurin 1/(5x+7) is 1/7-5/49 x+25/343 x^2-125/2401 x^3+625/16807 x^4+\ldots