What is the integral of x^5(4x^3-1)^4 ?

The integral of x^5(4x^3-1)^4 is (128x^{18})/9-(256x^{15}}{15}+8x^{12}-\frac{16x^9)/9+(x^6)/6+C

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integral of x^5(4x^3-1)^4

\int x^{5} (4 x^{3} - 1)^{4} d x

\frac{128 x ^{18}}{9} - \frac{256 x ^{15}}{15} + 8 x^{12} - \frac{16 x ^{9}}{9} + \frac{x ^{6}}{6} + C

Solution steps

\int x^{5} (4 x^{3} - 1)^{4} d x

Expand

x^{5} (4 x^{3} - 1)^{4} : 256 x^{17} - 256 x^{14} + 96 x^{11} - 16 x^{8} + x^{5}

= \int 256 x^{17} - 256 x^{14} + 96 x^{11} - 16 x^{8} + x^{5} 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^{17} d x - \int 256 x^{14} d x + \int 96 x^{11} d x - \int 16 x^{8} d x + \int x^{5} d x

\int 256 x^{17} d x = \frac{128 x ^{18}}{9}

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

\int 96 x^{11} d x = 8 x^{12}

\int 16 x^{8} d x = \frac{16 x ^{9}}{9}

\int x^{5} d x = \frac{x ^{6}}{6}

= \frac{128 x ^{18}}{9} - \frac{256 x ^{15}}{15} + 8 x^{12} - \frac{16 x ^{9}}{9} + \frac{x ^{6}}{6}

Add a constant to the solution

= \frac{128 x ^{18}}{9} - \frac{256 x ^{15}}{15} + 8 x^{12} - \frac{16 x ^{9}}{9} + \frac{x ^{6}}{6} + C

\int \frac{2 x + 1}{x ^{3} + x} d x

\int \frac{( x ^{2} + x )}{( x - 1 ) ( x ^{2} + 1 )} d x

\int \frac{x + 5 x ^{7}}{x ^{3}} d x

\frac{ln ( x )}{x ^{2}}

x ln (x)

What is the integral of x^5(4x^3-1)^4 ?
The integral of x^5(4x^3-1)^4 is (128x^{18})/9-(256x^{15}}{15}+8x^{12}-\frac{16x^9)/9+(x^6)/6+C