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

The integral of x(3x^3-2)^2 is (9x^8)/8-(12x^5)/5+2x^2+C

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

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

\frac{9 x ^{8}}{8} - \frac{12 x ^{5}}{5} + 2 x^{2} + C

Solution steps

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

Expand

x (3 x^{3} - 2)^{2} : 9 x^{7} - 12 x^{4} + 4 x

= \int 9 x^{7} - 12 x^{4} + 4 x 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^{7} d x - \int 12 x^{4} d x + \int 4 x d x

\int 9 x^{7} d x = \frac{9 x ^{8}}{8}

\int 12 x^{4} d x = \frac{12 x ^{5}}{5}

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

= \frac{9 x ^{8}}{8} - \frac{12 x ^{5}}{5} + 2 x^{2}

Add a constant to the solution

= \frac{9 x ^{8}}{8} - \frac{12 x ^{5}}{5} + 2 x^{2} + C

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What is the integral of x(3x^3-2)^2 ?
The integral of x(3x^3-2)^2 is (9x^8)/8-(12x^5)/5+2x^2+C