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

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

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

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

3 x^{4} + x^{3} - 21 x^{2}

y = x (25.5)

7 x^{2}

y = (arcsin (sec (x)))^{t a n h (x)}

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