What is the integral of x^4(1-x)^4 ?

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

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integral of x^4(1-x)^4

\int x^{4} (1 - x)^{4} d x

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

Solution steps

\int x^{4} (1 - x)^{4} d x

Expand

x^{4} (1 - x)^{4} : x^{4} - 4 x^{5} + 6 x^{6} - 4 x^{7} + x^{8}

= \int x^{4} - 4 x^{5} + 6 x^{6} - 4 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^{4} d x - \int 4 x^{5} d x + \int 6 x^{6} d x - \int 4 x^{7} d x + \int x^{8} d x

\int x^{4} d x = \frac{x ^{5}}{5}

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

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

\int 4 x^{7} d x = \frac{x ^{8}}{2}

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

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

Add a constant to the solution

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

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