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

The integral of (4x^2-2)^2x^2 is 16x^7-16x^5+4x^3-(96x^7)/7+(64x^5)/5-(8x^3)/3+C

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

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

16 x^{7} - 16 x^{5} + 4 x^{3} - \frac{96 x ^{7}}{7} + \frac{64 x ^{5}}{5} - \frac{8 x ^{3}}{3} + C

Solution steps

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

Factor

4 x^{2} - 2 : - (- 4 x^{2} + 2)

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

(- (- 4 x^{2} + 2))^{2} = (- 4 x^{2} + 2)^{2}

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

Apply Integration By Parts

= x^{3} (- 4 x^{2} + 2)^{2} - \int x (96 x^{5} - 64 x^{3} + 8 x) d x

\int x (96 x^{5} - 64 x^{3} + 8 x) d x = \frac{96 x ^{7}}{7} - \frac{64 x ^{5}}{5} + \frac{8 x ^{3}}{3}

= x^{3} (- 4 x^{2} + 2)^{2} - (\frac{96 x ^{7}}{7} - \frac{64 x ^{5}}{5} + \frac{8 x ^{3}}{3})

Simplify

x^{3} (- 4 x^{2} + 2)^{2} - (\frac{96 x ^{7}}{7} - \frac{64 x ^{5}}{5} + \frac{8 x ^{3}}{3}) : 16 x^{7} - 16 x^{5} + 4 x^{3} - \frac{96 x ^{7}}{7} + \frac{64 x ^{5}}{5} - \frac{8 x ^{3}}{3}

= 16 x^{7} - 16 x^{5} + 4 x^{3} - \frac{96 x ^{7}}{7} + \frac{64 x ^{5}}{5} - \frac{8 x ^{3}}{3}

Add a constant to the solution

= 16 x^{7} - 16 x^{5} + 4 x^{3} - \frac{96 x ^{7}}{7} + \frac{64 x ^{5}}{5} - \frac{8 x ^{3}}{3} + C

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What is the integral of (4x^2-2)^2x^2 ?
The integral of (4x^2-2)^2x^2 is 16x^7-16x^5+4x^3-(96x^7)/7+(64x^5)/5-(8x^3)/3+C