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

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

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

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

\frac{16 x ^{7}}{7} - 4 x^{4} + 4 x + C

Solution steps

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

Expand

(4 x^{3} - 2)^{2} : 16 x^{6} - 16 x^{3} + 4

= \int 16 x^{6} - 16 x^{3} + 4 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 16 x^{6} d x - \int 16 x^{3} d x + \int 4 d x

\int 16 x^{6} d x = \frac{16 x ^{7}}{7}

\int 16 x^{3} d x = 4 x^{4}

\int 4 d x = 4 x

= \frac{16 x ^{7}}{7} - 4 x^{4} + 4 x

Add a constant to the solution

= \frac{16 x ^{7}}{7} - 4 x^{4} + 4 x + C

\frac{d}{d x} (\frac{e ^{x}}{x ^{4} + 1})

\frac{d}{d x} (x^{6} - 1)

\int \frac{1}{x 4 x + 49} d x

(2 y^{2} + 2 y + 4 x^{2}) d x + (2 y x + x) d y = 0

\frac{\partial}{\partial x} (3 x^{2} y - 2 x y^{3})

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