What is the integral of (6x^3+3x^2) ?

The integral of (6x^3+3x^2) is (3x^4)/2+x^3+C

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integral of (6x^3+3x^2)

\int (6 x^{3} + 3 x^{2}) d x

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

Solution steps

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

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

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

Add a constant to the solution

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

d er i v a t i v e ln (sec (x))

x \to 0 lim (x^{2} cos (\frac{6}{x}))

\frac{d}{d x} ((1 + 4 x^{3}) (1 + 2 x^{2}))

\frac{\partial}{\partial s} (s + 2 t)

d er i v a t i v e 9 x^{2} - x^{3}

What is the integral of (6x^3+3x^2) ?
The integral of (6x^3+3x^2) is (3x^4)/2+x^3+C