What is the integral of 3v(v^2+4)^2 ?

The integral of 3v(v^2+4)^2 is 3((v^6)/6+2v^4+8v^2)+C

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integral of 3v(v^2+4)^2

\int 3 v (v^{2} + 4)^{2} d v

3 (\frac{v ^{6}}{6} + 2 v^{4} + 8 v^{2}) + C

Solution steps

\int 3 v (v^{2} + 4)^{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int v (v^{2} + 4)^{2} d v

Expand

v (v^{2} + 4)^{2} : v^{5} + 8 v^{3} + 16 v

= 3 \cdot \int v^{5} + 8 v^{3} + 16 v d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int v^{5} d v + \int 8 v^{3} d v + \int 16 v d v)

\int v^{5} d v = \frac{v ^{6}}{6}

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

\int 16 v d v = 8 v^{2}

= 3 (\frac{v ^{6}}{6} + 2 v^{4} + 8 v^{2})

Add a constant to the solution

= 3 (\frac{v ^{6}}{6} + 2 v^{4} + 8 v^{2}) + C

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What is the integral of 3v(v^2+4)^2 ?
The integral of 3v(v^2+4)^2 is 3((v^6)/6+2v^4+8v^2)+C