What is the integral of u(3u^2+1) ?

The integral of u(3u^2+1) is (3u^4)/4+(u^2)/2+C

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integral of u(3u^2+1)

\int u (3 u^{2} + 1) d u

\frac{3 u ^{4}}{4} + \frac{u ^{2}}{2} + C

Solution steps

\int u (3 u^{2} + 1) d u

Expand

u (3 u^{2} + 1) : 3 u^{3} + u

= \int 3 u^{3} + u 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^{3} d u + \int u d u

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

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

= \frac{3 u ^{4}}{4} + \frac{u ^{2}}{2}

Add a constant to the solution

= \frac{3 u ^{4}}{4} + \frac{u ^{2}}{2} + C

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What is the integral of u(3u^2+1) ?
The integral of u(3u^2+1) is (3u^4)/4+(u^2)/2+C