What is the integral of ((1-y)e^y)/(y^2) ?

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

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integral of ((1-y)e^y)/(y^2)

\int \frac{( 1 - y ) e ^{y}}{y ^{2}} d y

- \frac{e ^{y} ( 1 - y )}{y} - e^{y} + C

Solution steps

\int \frac{( 1 - y ) e ^{y}}{y ^{2}} d y

Apply Integration By Parts

: - \frac{e ^{y} ( 1 - y )}{y} - \int e^{y} d y

= - \frac{e ^{y} ( 1 - y )}{y} - \int e^{y} d y

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

= - \frac{e ^{y} ( 1 - y )}{y} - e^{y}

Add a constant to the solution

= - \frac{e ^{y} ( 1 - y )}{y} - e^{y} + C

y (8 x - 9 y) d x + 2 x (x - 3 y) d y = 0

\frac{d}{d x} (a x \cdot e^{b x})

f (x) = (11 x - 1)^{- \frac{1}{5}}, at x = 3

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

\frac{d}{d x} (ln ((2 x + 3)^{2}))

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