What is the integral of (4x^3-1)^2 ?

The integral of (4x^3-1)^2 is (16x^7)/7-2x^4+x+C

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integral of (4x^3-1)^2

\int (4 x^{3} - 1)^{2} d x

\frac{16 x ^{7}}{7} - 2 x^{4} + x + C

Solution steps

\int (4 x^{3} - 1)^{2} d x

Expand

(4 x^{3} - 1)^{2} : 16 x^{6} - 8 x^{3} + 1

= \int 16 x^{6} - 8 x^{3} + 1 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 8 x^{3} d x + \int 1 d x

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

\int 8 x^{3} d x = 2 x^{4}

\int 1 d x = x

= \frac{16 x ^{7}}{7} - 2 x^{4} + x

Add a constant to the solution

= \frac{16 x ^{7}}{7} - 2 x^{4} + x + C

f (x) = x^{\frac{1}{5}}

\frac{x + 1}{x ^{2} + 1}

(\frac{d y}{d x}) = (\frac{( x - y ^{2} )}{( 2 x y + y )})

\int \frac{x - arctan ( x )}{1 + x ^{2}} d x

x \to 2 - lim (8 x - x^{4})

What is the integral of (4x^3-1)^2 ?
The integral of (4x^3-1)^2 is (16x^7)/7-2x^4+x+C