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

The integral of (4x^2-2x)^2 is 1/320 (4x-1)^5-1/96 (4x-1)^3+1/16 (x-1/4)+C

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

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

\frac{1}{320} (4 x - 1)^{5} - \frac{1}{96} (4 x - 1)^{3} + \frac{1}{16} (x - \frac{1}{4}) + C

Solution steps

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

Complete the square

4 x^{2} - 2 x : 4 (x - \frac{1}{4})^{2} - \frac{1}{4}

= \int (4 (x - \frac{1}{4})^{2} - \frac{1}{4})^{2} d x

Apply u-substitution

: \int 16 u^{4} - 2 u^{2} + \frac{1}{16} d u

= \int 16 u^{4} - 2 u^{2} + \frac{1}{16} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 16 u^{4} d u - \int 2 u^{2} d u + \int \frac{1}{16} d u

\int 16 u^{4} d u = \frac{16 u ^{5}}{5}

\int 2 u^{2} d u = \frac{2 u ^{3}}{3}

\int \frac{1}{16} d u = \frac{1}{16} u

= \frac{16 u ^{5}}{5} - \frac{2 u ^{3}}{3} + \frac{1}{16} u

Substitute back

u = x - \frac{1}{4}

= \frac{16 ( x - \frac{1}{4} ) ^{5}}{5} - \frac{2 ( x - \frac{1}{4} ) ^{3}}{3} + \frac{1}{16} (x - \frac{1}{4})

Simplify

\frac{16 ( x - \frac{1}{4} ) ^{5}}{5} - \frac{2 ( x - \frac{1}{4} ) ^{3}}{3} + \frac{1}{16} (x - \frac{1}{4}) : \frac{1}{320} (4 x - 1)^{5} - \frac{1}{96} (4 x - 1)^{3} + \frac{1}{16} (x - \frac{1}{4})

= \frac{1}{320} (4 x - 1)^{5} - \frac{1}{96} (4 x - 1)^{3} + \frac{1}{16} (x - \frac{1}{4})

Add a constant to the solution

= \frac{1}{320} (4 x - 1)^{5} - \frac{1}{96} (4 x - 1)^{3} + \frac{1}{16} (x - \frac{1}{4}) + C

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What is the integral of (4x^2-2x)^2 ?
The integral of (4x^2-2x)^2 is 1/320 (4x-1)^5-1/96 (4x-1)^3+1/16 (x-1/4)+C