What is the integral of 8x(4x^2+4)^2 ?

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

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integral of 8x(4x^2+4)^2

\int 8 x (4 x^{2} + 4)^{2} d x

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

Solution steps

\int 8 x (4 x^{2} + 4)^{2} d x

Take the constant out:

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

= 8 \cdot \int x (4 x^{2} + 4)^{2} d x

Expand

x (4 x^{2} + 4)^{2} : 16 x^{5} + 32 x^{3} + 16 x

= 8 \cdot \int 16 x^{5} + 32 x^{3} + 16 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

= 8 (\int 16 x^{5} d x + \int 32 x^{3} d x + \int 16 x d x)

\int 16 x^{5} d x = \frac{8 x ^{6}}{3}

\int 32 x^{3} d x = 8 x^{4}

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

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

Add a constant to the solution

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

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