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

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

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

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

\frac{x ^{6}}{3} - 3 x^{4} + 9 x^{2} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

(x^{2} - 3)^{2} x : x^{5} - 6 x^{3} + 9 x

= 2 \cdot \int x^{5} - 6 x^{3} + 9 x 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

= 2 (\int x^{5} d x - \int 6 x^{3} d x + \int 9 x d x)

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

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

\int 9 x d x = \frac{9 x ^{2}}{2}

= 2 (\frac{x ^{6}}{6} - \frac{3 x ^{4}}{2} + \frac{9 x ^{2}}{2})

Simplify

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

= \frac{x ^{6}}{3} - 3 x^{4} + 9 x^{2}

Add a constant to the solution

= \frac{x ^{6}}{3} - 3 x^{4} + 9 x^{2} + C

x \to 0 + lim (\frac{1}{3 x})

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