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

The integral of 2x(3x-4)^4 is 27x^6-(864x^5)/5+432x^4-512x^3+256x^2+C

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

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

27 x^{6} - \frac{864 x ^{5}}{5} + 432 x^{4} - 512 x^{3} + 256 x^{2} + C

Solution steps

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

Take the constant out:

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

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

Expand

x (3 x - 4)^{4} : 81 x^{5} - 432 x^{4} + 864 x^{3} - 768 x^{2} + 256 x

= 2 \cdot \int 81 x^{5} - 432 x^{4} + 864 x^{3} - 768 x^{2} + 256 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

= 2 (\int 81 x^{5} d x - \int 432 x^{4} d x + \int 864 x^{3} d x - \int 768 x^{2} d x + \int 256 x d x)

\int 81 x^{5} d x = \frac{27 x ^{6}}{2}

\int 432 x^{4} d x = \frac{432 x ^{5}}{5}

\int 864 x^{3} d x = 216 x^{4}

\int 768 x^{2} d x = 256 x^{3}

\int 256 x d x = 128 x^{2}

= 2 (\frac{27 x ^{6}}{2} - \frac{432 x ^{5}}{5} + 216 x^{4} - 256 x^{3} + 128 x^{2})

Simplify

2 (\frac{27 x ^{6}}{2} - \frac{432 x ^{5}}{5} + 216 x^{4} - 256 x^{3} + 128 x^{2}) : 27 x^{6} - \frac{864 x ^{5}}{5} + 432 x^{4} - 512 x^{3} + 256 x^{2}

= 27 x^{6} - \frac{864 x ^{5}}{5} + 432 x^{4} - 512 x^{3} + 256 x^{2}

Add a constant to the solution

= 27 x^{6} - \frac{864 x ^{5}}{5} + 432 x^{4} - 512 x^{3} + 256 x^{2} + C

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What is the integral of 2x(3x-4)^4 ?
The integral of 2x(3x-4)^4 is 27x^6-(864x^5)/5+432x^4-512x^3+256x^2+C