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

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

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integral of x(x-2)^4

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

\frac{1}{2} x^{2} (x - 2)^{4} - 2 (\frac{x ^{6}}{6} - \frac{6 x ^{5}}{5} + 3 x^{4} - \frac{8 x ^{3}}{3}) + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{2} x^{2} (x - 2)^{4} - \int 2 x^{2} (x - 2)^{3} d x

= \frac{1}{2} x^{2} (x - 2)^{4} - \int 2 x^{2} (x - 2)^{3} d x

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

= \frac{1}{2} x^{2} (x - 2)^{4} - 2 (\frac{x ^{6}}{6} - \frac{6 x ^{5}}{5} + 3 x^{4} - \frac{8 x ^{3}}{3})

Add a constant to the solution

= \frac{1}{2} x^{2} (x - 2)^{4} - 2 (\frac{x ^{6}}{6} - \frac{6 x ^{5}}{5} + 3 x^{4} - \frac{8 x ^{3}}{3}) + C

n = 1 \sum \infty \frac{1}{n ^{2} + 300}

y^{^{'}} = 2 y (5 - y)

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

x \to 0 lim (\frac{x ^{2} - 5}{x - 2})

\frac{d ^{2}}{d x ^{2}} x y + 2 x^{2} = 5

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