What is the integral of x(1-x)^7 ?

The integral of x(1-x)^7 is (x^2)/2-(7x^3)/3+(21x^4)/4-7x^5+(35x^6)/6-3x^7+(7x^8)/8-(x^9)/9+C

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integral of x(1-x)^7

\int x (1 - x)^{7} d x

\frac{x ^{2}}{2} - \frac{7 x ^{3}}{3} + \frac{21 x ^{4}}{4} - 7 x^{5} + \frac{35 x ^{6}}{6} - 3 x^{7} + \frac{7 x ^{8}}{8} - \frac{x ^{9}}{9} + C

Solution steps

\int x (1 - x)^{7} d x

Expand

x (1 - x)^{7} : x - 7 x^{2} + 21 x^{3} - 35 x^{4} + 35 x^{5} - 21 x^{6} + 7 x^{7} - x^{8}

= \int x - 7 x^{2} + 21 x^{3} - 35 x^{4} + 35 x^{5} - 21 x^{6} + 7 x^{7} - x^{8} 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 d x - \int 7 x^{2} d x + \int 21 x^{3} d x - \int 35 x^{4} d x + \int 35 x^{5} d x - \int 21 x^{6} d x + \int 7 x^{7} d x - \int x^{8} d x

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

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

\int 21 x^{3} d x = \frac{21 x ^{4}}{4}

\int 35 x^{4} d x = 7 x^{5}

\int 35 x^{5} d x = \frac{35 x ^{6}}{6}

\int 21 x^{6} d x = 3 x^{7}

\int 7 x^{7} d x = \frac{7 x ^{8}}{8}

\int x^{8} d x = \frac{x ^{9}}{9}

= \frac{x ^{2}}{2} - \frac{7 x ^{3}}{3} + \frac{21 x ^{4}}{4} - 7 x^{5} + \frac{35 x ^{6}}{6} - 3 x^{7} + \frac{7 x ^{8}}{8} - \frac{x ^{9}}{9}

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{7 x ^{3}}{3} + \frac{21 x ^{4}}{4} - 7 x^{5} + \frac{35 x ^{6}}{6} - 3 x^{7} + \frac{7 x ^{8}}{8} - \frac{x ^{9}}{9} + C

f (x) = 3 x^{3} + 3 x^{2} - 11 x - 1

y^{^{''}} + 9 y = 3 csc^{2} (3 t)

\int sinh (2 x) d x

2 x^{\frac{1}{5}} (x^{3} - 8)

x \to 0 lim (\frac{x}{cos ( x )})

What is the integral of x(1-x)^7 ?
The integral of x(1-x)^7 is (x^2)/2-(7x^3)/3+(21x^4)/4-7x^5+(35x^6)/6-3x^7+(7x^8)/8-(x^9)/9+C