What is the integral of (3-x^2+4x^5-2x^{-3}) ?

The integral of (3-x^2+4x^5-2x^{-3}) is 3x-(x^3)/3+(2x^6)/3+1/(x^2)+C

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integral of (3-x^2+4x^5-2x^{-3})

\int (3 - x^{2} + 4 x^{5} - 2 x^{- 3}) d x

3 x - \frac{x ^{3}}{3} + \frac{2 x ^{6}}{3} + \frac{1}{x ^{2}} + C

Solution steps

\int 3 - x^{2} + 4 x^{5} - 2 x^{- 3} 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 3 d x - \int x^{2} d x + \int 4 x^{5} d x - \int 2 x^{- 3} d x

\int 3 d x = 3 x

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

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

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

= 3 x - \frac{x ^{3}}{3} + \frac{2 x ^{6}}{3} - (- \frac{1}{x ^{2}})

Simplify

= 3 x - \frac{x ^{3}}{3} + \frac{2 x ^{6}}{3} + \frac{1}{x ^{2}}

Add a constant to the solution

= 3 x - \frac{x ^{3}}{3} + \frac{2 x ^{6}}{3} + \frac{1}{x ^{2}} + C

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What is the integral of (3-x^2+4x^5-2x^{-3}) ?
The integral of (3-x^2+4x^5-2x^{-3}) is 3x-(x^3)/3+(2x^6)/3+1/(x^2)+C