What is the integral of 4x^3(x^4-8) ?

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

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integral of 4x^3(x^4-8)

\int 4 x^{3} (x^{4} - 8) d x

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

Solution steps

\int 4 x^{3} (x^{4} - 8) d x

Take the constant out:

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

= 4 \cdot \int x^{3} (x^{4} - 8) d x

Expand

x^{3} (x^{4} - 8) : x^{7} - 8 x^{3}

= 4 \cdot \int x^{7} - 8 x^{3} 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

= 4 (\int x^{7} d x - \int 8 x^{3} d x)

\int x^{7} d x = \frac{x ^{8}}{8}

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

= 4 (\frac{x ^{8}}{8} - 2 x^{4})

Add a constant to the solution

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

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