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

The integral of (x^2-4)^2(2x) is 2((x^6)/6-2x^4+8x^2)+C

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integral of (x^2-4)^2(2x)

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

2 (\frac{x ^{6}}{6} - 2 x^{4} + 8 x^{2}) + C

Solution steps

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

Expand

(x^{2} - 4)^{2} x : x^{5} - 8 x^{3} + 16 x

= 2 \cdot \int x^{5} - 8 x^{3} + 16 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 8 x^{3} d x + \int 16 x d x)

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

\int 8 x^{3} d x = 2 x^{4}

\int 16 x d x = 8 x^{2}

= 2 (\frac{x ^{6}}{6} - 2 x^{4} + 8 x^{2})

Add a constant to the solution

= 2 (\frac{x ^{6}}{6} - 2 x^{4} + 8 x^{2}) + C

\int 13 sin^{3} (x co) s^{2} x d x

\int \frac{arctan ( x )}{x} d x

f (x) = tan^{4} (x^{3})

x \to 0 lim (\frac{5}{x ^{2}})

ln (3 x^{3} + 6 x^{2} + 2 x + 6)

What is the integral of (x^2-4)^2(2x) ?
The integral of (x^2-4)^2(2x) is 2((x^6)/6-2x^4+8x^2)+C