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

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

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

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

\frac{x ^{5}}{5} - \frac{x ^{3}}{3} - 6 x + C

Solution steps

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

Expand

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

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

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

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

\int 6 d x = 6 x

= \frac{x ^{5}}{5} - \frac{x ^{3}}{3} - 6 x

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{x ^{3}}{3} - 6 x + C

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