What is the integral of 5x^3(x^4-3) ?

The integral of 5x^3(x^4-3) is 5((x^8)/8-(3x^4)/4)+C

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integral of 5x^3(x^4-3)

\int 5 x^{3} (x^{4} - 3) d x

5 (\frac{x ^{8}}{8} - \frac{3 x ^{4}}{4}) + C

Solution steps

\int 5 x^{3} (x^{4} - 3) d x

Take the constant out:

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

= 5 \cdot \int x^{3} (x^{4} - 3) d x

Expand

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

= 5 \cdot \int x^{7} - 3 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

= 5 (\int x^{7} d x - \int 3 x^{3} d x)

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

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

= 5 (\frac{x ^{8}}{8} - \frac{3 x ^{4}}{4})

Add a constant to the solution

= 5 (\frac{x ^{8}}{8} - \frac{3 x ^{4}}{4}) + C

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What is the integral of 5x^3(x^4-3) ?
The integral of 5x^3(x^4-3) is 5((x^8)/8-(3x^4)/4)+C