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

The integral of (3x^2-1)^2 is (9x^5)/5-2x^3+x+C

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

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

\frac{9 x ^{5}}{5} - 2 x^{3} + x + C

Solution steps

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

Expand

(3 x^{2} - 1)^{2} : 9 x^{4} - 6 x^{2} + 1

= \int 9 x^{4} - 6 x^{2} + 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 9 x^{4} d x - \int 6 x^{2} d x + \int 1 d x

\int 9 x^{4} d x = \frac{9 x ^{5}}{5}

\int 6 x^{2} d x = 2 x^{3}

\int 1 d x = x

= \frac{9 x ^{5}}{5} - 2 x^{3} + x

Add a constant to the solution

= \frac{9 x ^{5}}{5} - 2 x^{3} + x + C

What is the integral of (3x^2-1)^2 ?
The integral of (3x^2-1)^2 is (9x^5)/5-2x^3+x+C