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

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

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integral of (x^4-x)^2

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

\frac{x ^{9}}{9} - \frac{x ^{6}}{3} + \frac{x ^{3}}{3} + C

Solution steps

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

Expand

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

= \int x^{8} - 2 x^{5} + x^{2} 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 2 x^{5} d x + \int x^{2} d x

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

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

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

= \frac{x ^{9}}{9} - \frac{x ^{6}}{3} + \frac{x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{9}}{9} - \frac{x ^{6}}{3} + \frac{x ^{3}}{3} + C

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