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

The integral of (x^4-3)^2 is (x^9)/9-(6x^5)/5+9x+C

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integral of (x^4-3)^2

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

\frac{x ^{9}}{9} - \frac{6 x ^{5}}{5} + 9 x + C

Solution steps

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

Expand

(x^{4} - 3)^{2} : x^{8} - 6 x^{4} + 9

= \int x^{8} - 6 x^{4} + 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^{8} d x - \int 6 x^{4} d x + \int 9 d x

\int x^{8} d x = \frac{x ^{9}}{9}

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

\int 9 d x = 9 x

= \frac{x ^{9}}{9} - \frac{6 x ^{5}}{5} + 9 x

Add a constant to the solution

= \frac{x ^{9}}{9} - \frac{6 x ^{5}}{5} + 9 x + C

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What is the integral of (x^4-3)^2 ?
The integral of (x^4-3)^2 is (x^9)/9-(6x^5)/5+9x+C