What is the integral of x^2(2x-1)^5 ?

The integral of x^2(2x-1)^5 is 4x^8-(80x^7)/7+(40x^6)/3-8x^5+(5x^4)/2-(x^3)/3+C

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integral of x^2(2x-1)^5

\int x^{2} (2 x - 1)^{5} d x

4 x^{8} - \frac{80 x ^{7}}{7} + \frac{40 x ^{6}}{3} - 8 x^{5} + \frac{5 x ^{4}}{2} - \frac{x ^{3}}{3} + C

Solution steps

\int x^{2} (2 x - 1)^{5} d x

Expand

x^{2} (2 x - 1)^{5} : 32 x^{7} - 80 x^{6} + 80 x^{5} - 40 x^{4} + 10 x^{3} - x^{2}

= \int 32 x^{7} - 80 x^{6} + 80 x^{5} - 40 x^{4} + 10 x^{3} - 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 80 x^{6} d x + \int 80 x^{5} d x - \int 40 x^{4} d x + \int 10 x^{3} d x - \int x^{2} d x

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

\int 80 x^{6} d x = \frac{80 x ^{7}}{7}

\int 80 x^{5} d x = \frac{40 x ^{6}}{3}

\int 40 x^{4} d x = 8 x^{5}

\int 10 x^{3} d x = \frac{5 x ^{4}}{2}

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

= 4 x^{8} - \frac{80 x ^{7}}{7} + \frac{40 x ^{6}}{3} - 8 x^{5} + \frac{5 x ^{4}}{2} - \frac{x ^{3}}{3}

Add a constant to the solution

= 4 x^{8} - \frac{80 x ^{7}}{7} + \frac{40 x ^{6}}{3} - 8 x^{5} + \frac{5 x ^{4}}{2} - \frac{x ^{3}}{3} + C

\frac{d}{d x} (cos (x^{6}))

u^{4}

f (x) = (1 - x) (x^{2} - 3)^{2}, at x = 2

x \to 2 + lim ((2 - x)^{x - 2})

What is the integral of x^2(2x-1)^5 ?
The integral of x^2(2x-1)^5 is 4x^8-(80x^7)/7+(40x^6)/3-8x^5+(5x^4)/2-(x^3)/3+C