What is the integral of 64x^3(4x^2-1) ?

The integral of 64x^3(4x^2-1) is 64((2x^6)/3-(x^4)/4)+C

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integral of 64x^3(4x^2-1)

\int 64 x^{3} (4 x^{2} - 1) d x

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

Solution steps

\int 64 x^{3} (4 x^{2} - 1) d x

Take the constant out:

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

= 64 \cdot \int x^{3} (4 x^{2} - 1) d x

Expand

x^{3} (4 x^{2} - 1) : 4 x^{5} - x^{3}

= 64 \cdot \int 4 x^{5} - 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

= 64 (\int 4 x^{5} d x - \int x^{3} d x)

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

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

= 64 (\frac{2 x ^{6}}{3} - \frac{x ^{4}}{4})

Add a constant to the solution

= 64 (\frac{2 x ^{6}}{3} - \frac{x ^{4}}{4}) + C

x \to - 7 lim (h (x))

y = \frac{1}{18 x ^{2}}, y = x, y = \frac{x}{20}

\int_{- 3}^{- 1} \frac{2}{x} d x

x \to 5 lim (\frac{x + 3}{5 - x})

\frac{d}{d x} (x^{3} e^{- x} + ln (2 x))

What is the integral of 64x^3(4x^2-1) ?
The integral of 64x^3(4x^2-1) is 64((2x^6)/3-(x^4)/4)+C