What is the integral of (3x^2-4x^3) ?

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

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integral of (3x^2-4x^3)

\int (3 x^{2} - 4 x^{3}) d x

x^{3} - x^{4} + C

Solution steps

\int 3 x^{2} - 4 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^{2} d x - \int 4 x^{3} d x

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

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

= x^{3} - x^{4}

Add a constant to the solution

= x^{3} - x^{4} + C

x y^{^{'}} + 3 y^{^{'}} - (x + 3) cos (x) + y = 0

\int (1 - y)^{3} d y

\int y cos (4 y) d y

y = \frac{( 2 x ^{2} - 1 ) ^{3}}{( 5 x - 8 ) ^{2}}

\frac{\partial}{\partial x} (x x + 1)