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

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

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integral of (x^3-3)^2

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

\frac{x ^{7}}{7} - \frac{3 x ^{4}}{2} + 9 x + C

Solution steps

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

Expand

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

= \int x^{6} - 6 x^{3} + 9 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 6 x^{3} d x + \int 9 d x

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

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

\int 9 d x = 9 x

= \frac{x ^{7}}{7} - \frac{3 x ^{4}}{2} + 9 x

Add a constant to the solution

= \frac{x ^{7}}{7} - \frac{3 x ^{4}}{2} + 9 x + C

\int \frac{cos ( x )}{sin ^{8} ( x )} d x

x^{2} d x + 2 y d y = 0, y (0) = 2

y = \frac{7}{x} (1.7)

(\frac{x}{3} + 9)^{\frac{3}{2}}

\int_{0}^{- 1} \frac{x ^{3} + 8}{x + 2} d x

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