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

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

\frac{d}{d x} (\frac{x + 2}{x ^{2} + 4 x + 13})

\frac{\partial}{\partial y} ((x^{2} + y^{2} + z^{2})^{- \frac{1}{2}})

d er i v a t i v e f (x) = sin^{2} (tan (x))

\frac{d}{d t} (- \frac{1}{t + 1})

d er i v a t i v e f (x) = \frac{x}{a}

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