What is maclaurin (1+x)^{(-3)/2} ?

The answer to maclaurin (1+x)^{(-3)/2} is 1-3/2 x+15/8 x^2-35/16 x^3+315/128 x^4+\ldots

maclaurin (1+x)^{(-3)/2}

maclaurin

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

1 - \frac{3}{2} x + \frac{15}{8} x^{2} - \frac{35}{16} x^{3} + \frac{315}{128} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{- \frac{3}{2}}{1 !} x + \frac{\frac{15}{4}}{2 !} x^{2} + \frac{- \frac{105}{8}}{3 !} x^{3} + \frac{\frac{945}{16}}{4 !} x^{4} + \dots

Refine

= 1 - \frac{3}{2} x + \frac{15}{8} x^{2} - \frac{35}{16} x^{3} + \frac{315}{128} x^{4} + \dots

cos (sin (tan (3 x)))

x \to 0 lim ((e x - 1) cot (2 x))

\int \frac{x + 17}{x ^{2} + 16 x + 68} d x

\int \frac{( 4 x + 13 )}{( x ^{2} + 3 x - 10 )} d x

6 x^{2}, at x = 10

What is maclaurin (1+x)^{(-3)/2} ?
The answer to maclaurin (1+x)^{(-3)/2} is 1-3/2 x+15/8 x^2-35/16 x^3+315/128 x^4+\ldots