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

The integral of (x^2-4)^2 is (x^5)/5-(8x^3)/3+16x+C

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integral of (x^2-4)^2

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

\frac{x ^{5}}{5} - \frac{8 x ^{3}}{3} + 16 x + C

Solution steps

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

Expand

(x^{2} - 4)^{2} : x^{4} - 8 x^{2} + 16

= \int x^{4} - 8 x^{2} + 16 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^{4} d x - \int 8 x^{2} d x + \int 16 d x

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

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

\int 16 d x = 16 x

= \frac{x ^{5}}{5} - \frac{8 x ^{3}}{3} + 16 x

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{8 x ^{3}}{3} + 16 x + C

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What is the integral of (x^2-4)^2 ?
The integral of (x^2-4)^2 is (x^5)/5-(8x^3)/3+16x+C