What is the integral of 3x^3+3x-3 ?

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

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integral of 3x^3+3x-3

\int 3 x^{3} + 3 x - 3 d x

\frac{3 x ^{4}}{4} + \frac{3 x ^{2}}{2} - 3 x + C

Solution steps

\int 3 x^{3} + 3 x - 3 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 3 x^{3} d x + \int 3 x d x - \int 3 d x

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

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

\int 3 d x = 3 x

= \frac{3 x ^{4}}{4} + \frac{3 x ^{2}}{2} - 3 x

Add a constant to the solution

= \frac{3 x ^{4}}{4} + \frac{3 x ^{2}}{2} - 3 x + C

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