What is the integral of x^2(2x^2-4)^3 ?

The integral of x^2(2x^2-4)^3 is (8x^9)/9-(48x^7)/7+(96x^5)/5-(64x^3)/3+C

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integral of x^2(2x^2-4)^3

\int x^{2} (2 x^{2} - 4)^{3} d x

\frac{8 x ^{9}}{9} - \frac{48 x ^{7}}{7} + \frac{96 x ^{5}}{5} - \frac{64 x ^{3}}{3} + C

Solution steps

\int x^{2} (2 x^{2} - 4)^{3} d x

Expand

x^{2} (2 x^{2} - 4)^{3} : 8 x^{8} - 48 x^{6} + 96 x^{4} - 64 x^{2}

= \int 8 x^{8} - 48 x^{6} + 96 x^{4} - 64 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 8 x^{8} d x - \int 48 x^{6} d x + \int 96 x^{4} d x - \int 64 x^{2} d x

\int 8 x^{8} d x = \frac{8 x ^{9}}{9}

\int 48 x^{6} d x = \frac{48 x ^{7}}{7}

\int 96 x^{4} d x = \frac{96 x ^{5}}{5}

\int 64 x^{2} d x = \frac{64 x ^{3}}{3}

= \frac{8 x ^{9}}{9} - \frac{48 x ^{7}}{7} + \frac{96 x ^{5}}{5} - \frac{64 x ^{3}}{3}

Add a constant to the solution

= \frac{8 x ^{9}}{9} - \frac{48 x ^{7}}{7} + \frac{96 x ^{5}}{5} - \frac{64 x ^{3}}{3} + C

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What is the integral of x^2(2x^2-4)^3 ?
The integral of x^2(2x^2-4)^3 is (8x^9)/9-(48x^7)/7+(96x^5)/5-(64x^3)/3+C