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

The integral of (x^4-2x^2)(4x^2-4) is (4x^7)/7-(12x^5)/5+(8x^3)/3+C

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

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

\frac{4 x ^{7}}{7} - \frac{12 x ^{5}}{5} + \frac{8 x ^{3}}{3} + C

Solution steps

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

Expand

(x^{4} - 2 x^{2}) (4 x^{2} - 4) : 4 x^{6} - 12 x^{4} + 8 x^{2}

= \int 4 x^{6} - 12 x^{4} + 8 x^{2} 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 4 x^{6} d x - \int 12 x^{4} d x + \int 8 x^{2} d x

\int 4 x^{6} d x = \frac{4 x ^{7}}{7}

\int 12 x^{4} d x = \frac{12 x ^{5}}{5}

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

= \frac{4 x ^{7}}{7} - \frac{12 x ^{5}}{5} + \frac{8 x ^{3}}{3}

Add a constant to the solution

= \frac{4 x ^{7}}{7} - \frac{12 x ^{5}}{5} + \frac{8 x ^{3}}{3} + C

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What is the integral of (x^4-2x^2)(4x^2-4) ?
The integral of (x^4-2x^2)(4x^2-4) is (4x^7)/7-(12x^5)/5+(8x^3)/3+C