What is the integral of y^3+1.8y^2-2.4y ?

The integral of y^3+1.8y^2-2.4y is 0.25y^4+0.6y^3-1.2y^2+C

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integral of y^3+1.8y^2-2.4y

\int y^{3} + 1.8 y^{2} - 2.4 y d y

0.25 y^{4} + 0.6 y^{3} - 1.2 y^{2} + C

Solution steps

\int y^{3} + 1.8 y^{2} - 2.4 y d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int y^{3} d y + \int 1.8 y^{2} d y - \int 2.4 y d y

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

\int 1.8 y^{2} d y = 0.6 y^{3}

\int 2.4 y d y = 1.2 y^{2}

= \frac{y ^{4}}{4} + 0.6 y^{3} - 1.2 y^{2}

Simplify

= 0.25 y^{4} + 0.6 y^{3} - 1.2 y^{2}

Add a constant to the solution

= 0.25 y^{4} + 0.6 y^{3} - 1.2 y^{2} + C

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

\frac{\partial}{\partial y} (\frac{x y}{x + y + z})

\frac{\partial}{\partial y} (2 x y - y)

f (x) = \frac{7}{x}

\frac{sin ( x )}{cos ( x )}

What is the integral of y^3+1.8y^2-2.4y ?
The integral of y^3+1.8y^2-2.4y is 0.25y^4+0.6y^3-1.2y^2+C