What is the integral of x^2(x^3-1) ?

The integral of x^2(x^3-1) is (x^6)/6-(x^3)/3+C

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integral of x^2(x^3-1)

\int x^{2} (x^{3} - 1) d x

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

Solution steps

\int x^{2} (x^{3} - 1) d x

Expand

x^{2} (x^{3} - 1) : x^{5} - x^{2}

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

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

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

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

Add a constant to the solution

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

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What is the integral of x^2(x^3-1) ?
The integral of x^2(x^3-1) is (x^6)/6-(x^3)/3+C