What is the integral of x^4(1-x)^5 ?

The integral of x^4(1-x)^5 is (x^5)/5-(5x^6)/6+(10x^7)/7-(5x^8)/4+(5x^9)/9-(x^{10})/(10)+C

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integral of x^4(1-x)^5

integral

x^{4} (1 - x)^{5}

\frac{x ^{5}}{5} - \frac{5 x ^{6}}{6} + \frac{10 x ^{7}}{7} - \frac{5 x ^{8}}{4} + \frac{5 x ^{9}}{9} - \frac{x ^{10}}{10} + C

Solution steps

\int x^{4} (1 - x)^{5} d x

Expand

x^{4} (1 - x)^{5} : x^{4} - 5 x^{5} + 10 x^{6} - 10 x^{7} + 5 x^{8} - x^{9}

= \int x^{4} - 5 x^{5} + 10 x^{6} - 10 x^{7} + 5 x^{8} - x^{9} 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 5 x^{5} d x + \int 10 x^{6} d x - \int 10 x^{7} d x + \int 5 x^{8} d x - \int x^{9} d x

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

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

\int 10 x^{6} d x = \frac{10 x ^{7}}{7}

\int 10 x^{7} d x = \frac{5 x ^{8}}{4}

\int 5 x^{8} d x = \frac{5 x ^{9}}{9}

\int x^{9} d x = \frac{x ^{10}}{10}

= \frac{x ^{5}}{5} - \frac{5 x ^{6}}{6} + \frac{10 x ^{7}}{7} - \frac{5 x ^{8}}{4} + \frac{5 x ^{9}}{9} - \frac{x ^{10}}{10}

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{5 x ^{6}}{6} + \frac{10 x ^{7}}{7} - \frac{5 x ^{8}}{4} + \frac{5 x ^{9}}{9} - \frac{x ^{10}}{10} + C

\frac{x ^{3}}{( x ^{2} + x + 2 )}

(x^{2} + 1)^{3}

ln (x - 1)

\frac{1}{( a + x )}

(x + 1)^{\frac{1}{2}}

What is the integral of x^4(1-x)^5 ?
The integral of x^4(1-x)^5 is (x^5)/5-(5x^6)/6+(10x^7)/7-(5x^8)/4+(5x^9)/9-(x^{10})/(10)+C