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

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

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

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

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

Solution steps

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

Expand

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

= \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

= \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}

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

Add a constant to the solution

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

f (x) = 625 - 4.9 x^{2}, at x = 4

\frac{\partial}{\partial x} (x^{2} - 2 x y + y)

(0, 0), (8, 6)

n = 0 \sum \infty (1.67)^{n}

\frac{d y}{d x}, y^{9} = x^{8}

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