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

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

Popular >

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

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

\frac{v ^{6}}{6} + v^{4} + 2 v^{2} + C

Solution steps

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

Expand

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

= \int v^{5} + 4 v^{3} + 4 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

= \int v^{5} d v + \int 4 v^{3} d v + \int 4 v d v

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

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

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

= \frac{v ^{6}}{6} + v^{4} + 2 v^{2}

Add a constant to the solution

= \frac{v ^{6}}{6} + v^{4} + 2 v^{2} + C

y^{^{''}} - 6 y^{^{'}} + 25 y = 0

\frac{\partial}{\partial y} (x sin (6 x^{2} y))

\int 24 sec^{5} (x) d x

\frac{d y}{d x} = \frac{( y ^{2} - 1 )}{( x ^{2} - 1 )}, y (0) = 1

\frac{\partial}{\partial x} (4 x cos (x y))

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