What is the integral of-y^2e^y ?

The integral of-y^2e^y is -e^yy^2+2e^yy-2e^y+C

integral of-y^2e^y

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

- e^{y} y^{2} + 2 e^{y} y - 2 e^{y} + C

Solution steps

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

Take the constant out:

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

= - \int y^{2} e^{y} d y

Apply Integration By Parts

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

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

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

Simplify

- (y^{2} e^{y} - 2 (e^{y} y - e^{y})) : - e^{y} y^{2} + 2 e^{y} y - 2 e^{y}

= - e^{y} y^{2} + 2 e^{y} y - 2 e^{y}

Add a constant to the solution

= - e^{y} y^{2} + 2 e^{y} y - 2 e^{y} + C

\int (\frac{1}{4})^{x} d x

x \to - 3 lim (\frac{1 - 6 x}{3 + x})

\int_{0}^{1} \frac{x ^{2}}{x ^{3} + 1} d x

\frac{d}{d x} (\frac{sin ( \frac{1}{x} ) x ^{3}}{ln ( x )})

a re a 1 - x^{2}, x^{6} - x + 1