What is the integral of (x^{-3}-6x^{-2}+10) ?

The integral of (x^{-3}-6x^{-2}+10) is -1/(2x^2)+6/x+10x+C

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integral of (x^{-3}-6x^{-2}+10)

\int (x^{- 3} - 6 x^{- 2} + 10) d x

- \frac{1}{2 x ^{2}} + \frac{6}{x} + 10 x + C

Solution steps

\int x^{- 3} - 6 x^{- 2} + 10 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^{- 3} d x - \int 6 x^{- 2} d x + \int 10 d x

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

\int 6 x^{- 2} d x = - \frac{6}{x}

\int 10 d x = 10 x

= - \frac{1}{2 x ^{2}} - (- \frac{6}{x}) + 10 x

Simplify

= - \frac{1}{2 x ^{2}} + \frac{6}{x} + 10 x

Add a constant to the solution

= - \frac{1}{2 x ^{2}} + \frac{6}{x} + 10 x + C

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What is the integral of (x^{-3}-6x^{-2}+10) ?
The integral of (x^{-3}-6x^{-2}+10) is -1/(2x^2)+6/x+10x+C