How do you integrate $f (x) = x^{3} - 3 x - 2 x^{- 4}$ $f(x)=x^3-3x-2x^-4$ using the power rule?

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Answer 1 · 2016-11-18T16:01:53

$\frac{1}{4} x^{4} - \frac{3}{2} x^{2} + \frac{2}{3} x^{- 3} + C$ $1/4x^4-3/2x^2+2/3x^-3+C$

for polynomials

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C, n \neq - 1$ $intx^ndx=x^(n+1)/(n+1)+C,n!=-1$

$\int (x^{3} - 3 x - 2 x^{- 4}) d x$ $int(x^3-3x-2x^(-4))dx$

$= \frac{x^{3 + 1}}{3 + 1} - 3 \frac{x^{1 + 1}}{1 + 1} - 2 \frac{x^{- 4 + 1}}{- 4 + 1} + C$ $=x^(3+1)/(3+1)-3x^(1+1)/(1+1)-2x^(-4+1)/(-4+1)+C$

$= \frac{x^{4}}{4} - \frac{3 x^{2}}{2} - \frac{2 x^{- 3}}{- 3} + C$ $=x^4/4-(3x^2)/2-(2x^-3)/-3+C$

tidying up

$\frac{1}{4} x^{4} - \frac{3}{2} x^{2} + \frac{2}{3} x^{- 3} + C$ $1/4x^4-3/2x^2+2/3x^-3+C$

How do you integrate f(x)=x3−3x−2x−4f(x)=x^3-3x-2x^-4 using the power rule?