What is the integral of 2pix(x^2-1)^2 ?

The integral of 2pix(x^2-1)^2 is (pix^6)/3-pix^4+pix^2+C

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integral of 2pix(x^2-1)^2

\int 2 π x (x^{2} - 1)^{2} d x

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

Solution steps

\int 2 π x (x^{2} - 1)^{2} d x

Take the constant out:

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

= 2 π \cdot \int x (x^{2} - 1)^{2} d x

Expand

x (x^{2} - 1)^{2} : x^{5} - 2 x^{3} + x

= 2 π \cdot \int x^{5} - 2 x^{3} + 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 π (\int x^{5} d x - \int 2 x^{3} d x + \int x d x)

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int_{4}^{9} \frac{4}{2 x} d x

f (x) = 7 arctan (7 x), x = 1

n = 0 \sum \infty n^{2} (\frac{1}{3})^{n}

\int e^{- s t} t^{5} d t

f^{^{'}} (x) = \frac{1}{3 x ^{\frac{2}{3}}}

What is the integral of 2pix(x^2-1)^2 ?
The integral of 2pix(x^2-1)^2 is (pix^6)/3-pix^4+pix^2+C