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

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

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

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

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

Solution steps

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

Expand

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

= \int 4 x^{6} - 4 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 4 x^{6} d x - \int 4 x^{3} d x + \int 1 d x

\int 4 x^{6} d x = \frac{4 x ^{7}}{7}

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

\int 1 d x = x

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

Add a constant to the solution

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

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