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

The integral of 64x^3(4x^2-1)^3 is 64((32x^{10})/5-6x^8+2x^6-(x^4)/4)+C

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

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

64 (\frac{32 x ^{10}}{5} - 6 x^{8} + 2 x^{6} - \frac{x ^{4}}{4}) + C

Solution steps

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

Expand

x^{3} (4 x^{2} - 1)^{3} : 64 x^{9} - 48 x^{7} + 12 x^{5} - x^{3}

= 64 \cdot \int 64 x^{9} - 48 x^{7} + 12 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 64 x^{9} d x - \int 48 x^{7} d x + \int 12 x^{5} d x - \int x^{3} d x)

\int 64 x^{9} d x = \frac{32 x ^{10}}{5}

\int 48 x^{7} d x = 6 x^{8}

\int 12 x^{5} d x = 2 x^{6}

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

= 64 (\frac{32 x ^{10}}{5} - 6 x^{8} + 2 x^{6} - \frac{x ^{4}}{4})

Add a constant to the solution

= 64 (\frac{32 x ^{10}}{5} - 6 x^{8} + 2 x^{6} - \frac{x ^{4}}{4}) + C

\frac{d}{d x} (x^{\frac{4}{5}} (x - 8)^{2})

\int_{- 1}^{6} x - 4 x^{2} d x

y^{^{''}} - (\frac{2}{t}) y^{^{'}} + (\frac{2}{t ^{2}}) y = 0

\int (ln (y))^{2} y d y

\frac{x ^{2} + 4}{x ^{2} - 4}

What is the integral of 64x^3(4x^2-1)^3 ?
The integral of 64x^3(4x^2-1)^3 is 64((32x^{10})/5-6x^8+2x^6-(x^4)/4)+C