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

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

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integral of 2x(x^2+2)^2

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

2 (\frac{x ^{6}}{6} + x^{4} + 2 x^{2}) + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int x (x^{2} + 2)^{2} d x

Expand

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

= 2 \cdot \int x^{5} + 4 x^{3} + 4 x d x

Apply the Sum Rule:

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

= 2 (\int x^{5} d x + \int 4 x^{3} d x + \int 4 x d x)

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

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

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

= 2 (\frac{x ^{6}}{6} + x^{4} + 2 x^{2})

Add a constant to the solution

= 2 (\frac{x ^{6}}{6} + x^{4} + 2 x^{2}) + C

\int 3 cos^{2} (x) d x

f (x) = x sin (x) + cos (x)

\frac{\partial}{\partial x} (sin (3 x + yz))

\int_{3}^{\infty} \frac{8}{x ^{2} - x} d x

F (s) = \frac{6}{s ^{2} + 4}

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