What is the integral of 4y^3e^y ?

The integral of 4y^3e^y is 4(y^3e^y-3(y^2e^y-2(e^yy-e^y)))+C

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integral of 4y^3e^y

\int 4 y^{3} e^{y} d y

4 (y^{3} e^{y} - 3 (y^{2} e^{y} - 2 (e^{y} y - e^{y}))) + C

Solution steps

\int 4 y^{3} e^{y} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int y^{3} e^{y} d y

Apply Integration By Parts

: y^{3} e^{y} - \int 3 y^{2} e^{y} d y

= 4 (y^{3} e^{y} - \int 3 y^{2} e^{y} d y)

\int 3 y^{2} e^{y} d y = 3 (y^{2} e^{y} - 2 (e^{y} y - e^{y}))

= 4 (y^{3} e^{y} - 3 (y^{2} e^{y} - 2 (e^{y} y - e^{y})))

Add a constant to the solution

= 4 (y^{3} e^{y} - 3 (y^{2} e^{y} - 2 (e^{y} y - e^{y}))) + C

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