What is the integral of x^2(2x-4)^5 ?

The integral of x^2(2x-4)^5 is 4x^8-(320x^7)/7+(640x^6)/3-512x^5+640x^4-(1024x^3)/3+C

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integral of x^2(2x-4)^5

\int x^{2} (2 x - 4)^{5} d x

4 x^{8} - \frac{320 x ^{7}}{7} + \frac{640 x ^{6}}{3} - 512 x^{5} + 640 x^{4} - \frac{1024 x ^{3}}{3} + C

Solution steps

\int x^{2} (2 x - 4)^{5} d x

Expand

x^{2} (2 x - 4)^{5} : 32 x^{7} - 320 x^{6} + 1280 x^{5} - 2560 x^{4} + 2560 x^{3} - 1024 x^{2}

= \int 32 x^{7} - 320 x^{6} + 1280 x^{5} - 2560 x^{4} + 2560 x^{3} - 1024 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 32 x^{7} d x - \int 320 x^{6} d x + \int 1280 x^{5} d x - \int 2560 x^{4} d x + \int 2560 x^{3} d x - \int 1024 x^{2} d x

\int 32 x^{7} d x = 4 x^{8}

\int 320 x^{6} d x = \frac{320 x ^{7}}{7}

\int 1280 x^{5} d x = \frac{640 x ^{6}}{3}

\int 2560 x^{4} d x = 512 x^{5}

\int 2560 x^{3} d x = 640 x^{4}

\int 1024 x^{2} d x = \frac{1024 x ^{3}}{3}

= 4 x^{8} - \frac{320 x ^{7}}{7} + \frac{640 x ^{6}}{3} - 512 x^{5} + 640 x^{4} - \frac{1024 x ^{3}}{3}

Add a constant to the solution

= 4 x^{8} - \frac{320 x ^{7}}{7} + \frac{640 x ^{6}}{3} - 512 x^{5} + 640 x^{4} - \frac{1024 x ^{3}}{3} + C

\frac{d}{d x} (\frac{( ln ( x ) )}{x})

\int_{0}^{\frac{3}{4}} sin^{2} (π x) d x

\frac{\partial}{\partial y} (2 x y - e^{2 y})

\frac{\partial}{\partial x} ((e^{x} + e^{y})^{9})

f (x) = 3 e^{4 x}

What is the integral of x^2(2x-4)^5 ?
The integral of x^2(2x-4)^5 is 4x^8-(320x^7)/7+(640x^6)/3-512x^5+640x^4-(1024x^3)/3+C