What is the integral of 60x^2(1-4x^3)^4 ?

The integral of 60x^2(1-4x^3)^4 is 60((x^3)/3-(8x^6)/3+(32x^9)/3-(64x^{12})/3+(256x^{15})/(15))+C

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integral of 60x^2(1-4x^3)^4

\int 60 x^{2} (1 - 4 x^{3})^{4} d x

60 (\frac{x ^{3}}{3} - \frac{8 x ^{6}}{3} + \frac{32 x ^{9}}{3} - \frac{64 x ^{12}}{3} + \frac{256 x ^{15}}{15}) + C

Solution steps

\int 60 x^{2} (1 - 4 x^{3})^{4} d x

Take the constant out:

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

= 60 \cdot \int x^{2} (1 - 4 x^{3})^{4} d x

Factor

1 - 4 x^{3} : - (- 1 + 4 x^{3})

= 60 \cdot \int x^{2} (- (- 1 + 4 x^{3}))^{4} d x

(- (- 1 + 4 x^{3}))^{4} = (- 1 + 4 x^{3})^{4}

= 60 \cdot \int x^{2} (- 1 + 4 x^{3})^{4} d x

Expand

x^{2} (- 1 + 4 x^{3})^{4} : x^{2} - 16 x^{5} + 96 x^{8} - 256 x^{11} + 256 x^{14}

= 60 \cdot \int x^{2} - 16 x^{5} + 96 x^{8} - 256 x^{11} + 256 x^{14} 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

= 60 (\int x^{2} d x - \int 16 x^{5} d x + \int 96 x^{8} d x - \int 256 x^{11} d x + \int 256 x^{14} d x)

\int x^{2} d x = \frac{x ^{3}}{3}

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

\int 96 x^{8} d x = \frac{32 x ^{9}}{3}

\int 256 x^{11} d x = \frac{64 x ^{12}}{3}

\int 256 x^{14} d x = \frac{256 x ^{15}}{15}

= 60 (\frac{x ^{3}}{3} - \frac{8 x ^{6}}{3} + \frac{32 x ^{9}}{3} - \frac{64 x ^{12}}{3} + \frac{256 x ^{15}}{15})

Add a constant to the solution

= 60 (\frac{x ^{3}}{3} - \frac{8 x ^{6}}{3} + \frac{32 x ^{9}}{3} - \frac{64 x ^{12}}{3} + \frac{256 x ^{15}}{15}) + C

f (x) = x^{\frac{5}{6}}

y = 10 - 3 x - x^{2}

\frac{d y}{d x} + y cos (x) = 4 cos (x), y (0) = 6

\int 3 x 4 - x^{2} d x

What is the integral of 60x^2(1-4x^3)^4 ?
The integral of 60x^2(1-4x^3)^4 is 60((x^3)/3-(8x^6)/3+(32x^9)/3-(64x^{12})/3+(256x^{15})/(15))+C