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

The integral of x(x-4)^5 is (x^7)/7-(10x^6)/3+32x^5-160x^4+(1280x^3)/3-512x^2+C

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integral of x(x-4)^5

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

\frac{x ^{7}}{7} - \frac{10 x ^{6}}{3} + 32 x^{5} - 160 x^{4} + \frac{1280 x ^{3}}{3} - 512 x^{2} + C

Solution steps

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

Expand

x (x - 4)^{5} : x^{6} - 20 x^{5} + 160 x^{4} - 640 x^{3} + 1280 x^{2} - 1024 x

= \int x^{6} - 20 x^{5} + 160 x^{4} - 640 x^{3} + 1280 x^{2} - 1024 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

= \int x^{6} d x - \int 20 x^{5} d x + \int 160 x^{4} d x - \int 640 x^{3} d x + \int 1280 x^{2} d x - \int 1024 x d x

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

\int 20 x^{5} d x = \frac{10 x ^{6}}{3}

\int 160 x^{4} d x = 32 x^{5}

\int 640 x^{3} d x = 160 x^{4}

\int 1280 x^{2} d x = \frac{1280 x ^{3}}{3}

\int 1024 x d x = 512 x^{2}

= \frac{x ^{7}}{7} - \frac{10 x ^{6}}{3} + 32 x^{5} - 160 x^{4} + \frac{1280 x ^{3}}{3} - 512 x^{2}

Add a constant to the solution

= \frac{x ^{7}}{7} - \frac{10 x ^{6}}{3} + 32 x^{5} - 160 x^{4} + \frac{1280 x ^{3}}{3} - 512 x^{2} + C

n \to \infty lim (3^{n})

\frac{d}{d x} ((x + 2)^{3} (2 x - 1)^{5})

\int_{0}^{x} (sin^{2} (t)) d t

\frac{d y}{d x} = - 0.13 \cdot (100 - y)

n = 0 \sum \infty \frac{n}{( n + 1 )}

What is the integral of x(x-4)^5 ?
The integral of x(x-4)^5 is (x^7)/7-(10x^6)/3+32x^5-160x^4+(1280x^3)/3-512x^2+C