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

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

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integral of (x^2-3)^2

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

\frac{x ^{5}}{5} - 2 x^{3} + 9 x + C

Solution steps

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

Expand

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

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

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

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

\int 9 d x = 9 x

= \frac{x ^{5}}{5} - 2 x^{3} + 9 x

Add a constant to the solution

= \frac{x ^{5}}{5} - 2 x^{3} + 9 x + C

6 8 x^{2} + 9

\frac{d}{d x} (\frac{2 x - 1}{x + 3})

x \to 3 lim (- 2)

\int - 8 x e^{x^{2}} d x

\int sin^{5} (x) cos^{3} (x) d x

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