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

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

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integral of (x^2-x)^4

integral

(x^{2} - x)^{4}

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

Solution steps

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

Factor

x^{2} - x : - (- x^{2} + x)

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

Simplify

(- (- x^{2} + x))^{4} : (- x^{2} + x)^{4}

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

Expand

(- x^{2} + x)^{4} : x^{8} - 4 x^{7} + 6 x^{6} - 4 x^{5} + x^{4}

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

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

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

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

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

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

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

Add a constant to the solution

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

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