What is the integral of (3x^5-x^3+6) ?

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

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integral of (3x^5-x^3+6)

\int (3 x^{5} - x^{3} + 6) d x

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

Solution steps

\int 3 x^{5} - x^{3} + 6 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 x^{5} d x - \int x^{3} d x + \int 6 d x

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

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

\int 6 d x = 6 x

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

Add a constant to the solution

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

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