What is the integral of (x^2-4)^3 ?

The integral of (x^2-4)^3 is (x^7)/7-(12x^5)/5+16x^3-64x+C

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integral of (x^2-4)^3

\int (x^{2} - 4)^{3} d x

\frac{x ^{7}}{7} - \frac{12 x ^{5}}{5} + 16 x^{3} - 64 x + C

Solution steps

\int (x^{2} - 4)^{3} d x

Expand

(x^{2} - 4)^{3} : x^{6} - 12 x^{4} + 48 x^{2} - 64

= \int x^{6} - 12 x^{4} + 48 x^{2} - 64 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 12 x^{4} d x + \int 48 x^{2} d x - \int 64 d x

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

\int 12 x^{4} d x = \frac{12 x ^{5}}{5}

\int 48 x^{2} d x = 16 x^{3}

\int 64 d x = 64 x

= \frac{x ^{7}}{7} - \frac{12 x ^{5}}{5} + 16 x^{3} - 64 x

Add a constant to the solution

= \frac{x ^{7}}{7} - \frac{12 x ^{5}}{5} + 16 x^{3} - 64 x + C

What is the integral of (x^2-4)^3 ?
The integral of (x^2-4)^3 is (x^7)/7-(12x^5)/5+16x^3-64x+C