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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

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

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

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

Add a constant to the solution

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

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