What is the integral of (3y^3-2y^2+(e^y)/6) ?

The integral of (3y^3-2y^2+(e^y)/6) is (3y^4)/4-(2y^3)/3+1/6 e^y+C

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integral of (3y^3-2y^2+(e^y)/6)

\int (3 y^{3} - 2 y^{2} + \frac{e ^{y}}{6}) d y

\frac{3 y ^{4}}{4} - \frac{2 y ^{3}}{3} + \frac{1}{6} e^{y} + C

Solution steps

\int 3 y^{3} - 2 y^{2} + \frac{e ^{y}}{6} 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 3 y^{3} d y - \int 2 y^{2} d y + \int \frac{e ^{y}}{6} d y

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

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

\int \frac{e ^{y}}{6} d y = \frac{1}{6} e^{y}

= \frac{3 y ^{4}}{4} - \frac{2 y ^{3}}{3} + \frac{1}{6} e^{y}

Add a constant to the solution

= \frac{3 y ^{4}}{4} - \frac{2 y ^{3}}{3} + \frac{1}{6} e^{y} + C

\frac{d}{d x} (\frac{- 2 x}{- 4 x ^{4}})

\int_{0}^{2} \frac{1}{x ^{2} + 4} d x

\frac{\partial}{\partial x} (x^{4} y^{3} + 8 x^{2} y)

\frac{d}{d x} (\frac{x - 5}{5})

81 - y^{2}

What is the integral of (3y^3-2y^2+(e^y)/6) ?
The integral of (3y^3-2y^2+(e^y)/6) is (3y^4)/4-(2y^3)/3+1/6 e^y+C