What is the integral of (8x^3-6x^{-2}) ?

The integral of (8x^3-6x^{-2}) is 2x^4+6/x+C

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integral of (8x^3-6x^{-2})

\int (8 x^{3} - 6 x^{- 2}) d x

2 x^{4} + \frac{6}{x} + C

Solution steps

\int 8 x^{3} - 6 x^{- 2} d x

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

= \int 8 x^{3} - \frac{6}{x ^{2}} 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 8 x^{3} d x - \int \frac{6}{x ^{2}} d x

\int 8 x^{3} d x = 2 x^{4}

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

= 2 x^{4} - (- \frac{6}{x})

Simplify

= 2 x^{4} + \frac{6}{x}

Add a constant to the solution

= 2 x^{4} + \frac{6}{x} + C

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What is the integral of (8x^3-6x^{-2}) ?
The integral of (8x^3-6x^{-2}) is 2x^4+6/x+C