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

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

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integral of (x^3-1)^2

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

\frac{x ^{7}}{7} - \frac{x ^{4}}{2} + x + C

Solution steps

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

Expand

(x^{3} - 1)^{2} : x^{6} - 2 x^{3} + 1

= \int x^{6} - 2 x^{3} + 1 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^{6} d x - \int 2 x^{3} d x + \int 1 d x

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

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

\int 1 d x = x

= \frac{x ^{7}}{7} - \frac{x ^{4}}{2} + x

Add a constant to the solution

= \frac{x ^{7}}{7} - \frac{x ^{4}}{2} + x + C

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