What is the integral of (x^{-4}+2x^3-4x) ?

The integral of (x^{-4}+2x^3-4x) is -1/(3x^3)+(x^4)/2-2x^2+C

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integral of (x^{-4}+2x^3-4x)

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

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

Solution steps

\int x^{- 4} + 2 x^{3} - 4 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{- 4} d x + \int 2 x^{3} d x - \int 4 x d x

\int x^{- 4} d x = - \frac{1}{3 x ^{3}}

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int 4 x d x = 2 x^{2}

= - \frac{1}{3 x ^{3}} + \frac{x ^{4}}{2} - 2 x^{2}

Add a constant to the solution

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

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What is the integral of (x^{-4}+2x^3-4x) ?
The integral of (x^{-4}+2x^3-4x) is -1/(3x^3)+(x^4)/2-2x^2+C