What is the integral of 8x(2x-1)^4 ?

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

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integral of 8x(2x-1)^4

\int 8 x (2 x - 1)^{4} d x

8 (\frac{8 x ^{6}}{3} - \frac{32 x ^{5}}{5} + 6 x^{4} - \frac{8 x ^{3}}{3} + \frac{x ^{2}}{2}) + C

Solution steps

\int 8 x (2 x - 1)^{4} d x

Take the constant out:

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

= 8 \cdot \int x (2 x - 1)^{4} d x

Expand

x (2 x - 1)^{4} : 16 x^{5} - 32 x^{4} + 24 x^{3} - 8 x^{2} + x

= 8 \cdot \int 16 x^{5} - 32 x^{4} + 24 x^{3} - 8 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

= 8 (\int 16 x^{5} d x - \int 32 x^{4} d x + \int 24 x^{3} d x - \int 8 x^{2} d x + \int x d x)

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

\int 32 x^{4} d x = \frac{32 x ^{5}}{5}

\int 24 x^{3} d x = 6 x^{4}

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

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

= 8 (\frac{8 x ^{6}}{3} - \frac{32 x ^{5}}{5} + 6 x^{4} - \frac{8 x ^{3}}{3} + \frac{x ^{2}}{2})

Add a constant to the solution

= 8 (\frac{8 x ^{6}}{3} - \frac{32 x ^{5}}{5} + 6 x^{4} - \frac{8 x ^{3}}{3} + \frac{x ^{2}}{2}) + C

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