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

The integral of x(4x-1)^4 is (128x^6)/3-(256x^5)/5+24x^4-(16x^3)/3+(x^2)/2+C

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integral of x(4x-1)^4

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

\frac{128 x ^{6}}{3} - \frac{256 x ^{5}}{5} + 24 x^{4} - \frac{16 x ^{3}}{3} + \frac{x ^{2}}{2} + C

Solution steps

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

Expand

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

= \int 256 x^{5} - 256 x^{4} + 96 x^{3} - 16 x^{2} + x 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^{5} d x - \int 256 x^{4} d x + \int 96 x^{3} d x - \int 16 x^{2} d x + \int x d x

\int 256 x^{5} d x = \frac{128 x ^{6}}{3}

\int 256 x^{4} d x = \frac{256 x ^{5}}{5}

\int 96 x^{3} d x = 24 x^{4}

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

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

= \frac{128 x ^{6}}{3} - \frac{256 x ^{5}}{5} + 24 x^{4} - \frac{16 x ^{3}}{3} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{128 x ^{6}}{3} - \frac{256 x ^{5}}{5} + 24 x^{4} - \frac{16 x ^{3}}{3} + \frac{x ^{2}}{2} + C

What is the integral of x(4x-1)^4 ?
The integral of x(4x-1)^4 is (128x^6)/3-(256x^5)/5+24x^4-(16x^3)/3+(x^2)/2+C