What is the integral of 7v(v^2+4)^2 ?

The integral of 7v(v^2+4)^2 is 7((v^6)/6+2v^4+8v^2)+C

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integral of 7v(v^2+4)^2

\int 7 v (v^{2} + 4)^{2} d v

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

Solution steps

\int 7 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

= 7 \cdot \int v (v^{2} + 4)^{2} d v

Expand

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

= 7 \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

= 7 (\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}

= 7 (\frac{v ^{6}}{6} + 2 v^{4} + 8 v^{2})

Add a constant to the solution

= 7 (\frac{v ^{6}}{6} + 2 v^{4} + 8 v^{2}) + C

\int 2 + 2 sin (x) d x

\int 2 + \frac{1}{x} d x

f (x) = \frac{- 8 x}{x ^{2} + 1}, at x = 1

\frac{\partial}{\partial x} (2 x^{2} y + 4)

(0, 2), (1, 0)

What is the integral of 7v(v^2+4)^2 ?
The integral of 7v(v^2+4)^2 is 7((v^6)/6+2v^4+8v^2)+C