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

The integral of (x^2-3)^3 is (x^7)/7-(9x^5)/5+9x^3-27x+C

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integral of (x^2-3)^3

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

\frac{x ^{7}}{7} - \frac{9 x ^{5}}{5} + 9 x^{3} - 27 x + C

Solution steps

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

Expand

(x^{2} - 3)^{3} : x^{6} - 9 x^{4} + 27 x^{2} - 27

= \int x^{6} - 9 x^{4} + 27 x^{2} - 27 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 9 x^{4} d x + \int 27 x^{2} d x - \int 27 d x

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

\int 9 x^{4} d x = \frac{9 x ^{5}}{5}

\int 27 x^{2} d x = 9 x^{3}

\int 27 d x = 27 x

= \frac{x ^{7}}{7} - \frac{9 x ^{5}}{5} + 9 x^{3} - 27 x

Add a constant to the solution

= \frac{x ^{7}}{7} - \frac{9 x ^{5}}{5} + 9 x^{3} - 27 x + C

What is the integral of (x^2-3)^3 ?
The integral of (x^2-3)^3 is (x^7)/7-(9x^5)/5+9x^3-27x+C