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

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

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integral of (2x^2-3)^2

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

\frac{4 x ^{5}}{5} - 4 x^{3} + 9 x + C

Solution steps

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

Expand

(2 x^{2} - 3)^{2} : 4 x^{4} - 12 x^{2} + 9

= \int 4 x^{4} - 12 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 4 x^{4} d x - \int 12 x^{2} d x + \int 9 d x

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

\int 12 x^{2} d x = 4 x^{3}

\int 9 d x = 9 x

= \frac{4 x ^{5}}{5} - 4 x^{3} + 9 x

Add a constant to the solution

= \frac{4 x ^{5}}{5} - 4 x^{3} + 9 x + C

x \to 2 lim (\frac{x}{x - 2})

x \to \infty lim (x^{3})

cos (x) + i sin (x)

\int e^{2 x} x d x

\int 9 x^{2} - 24 x - 36 d x

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