What is the integral of (3u^5-2u^3) ?

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

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integral of (3u^5-2u^3)

\int (3 u^{5} - 2 u^{3}) d u

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

Solution steps

\int 3 u^{5} - 2 u^{3} d u

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 u^{5} d u - \int 2 u^{3} d u

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

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

= \frac{u ^{6}}{2} - \frac{u ^{4}}{2}

Add a constant to the solution

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

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