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

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

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

\int \frac{y - 1}{e ^{y}} d y

- \frac{y}{e ^{y}} + C

Solution steps

\int \frac{y - 1}{e ^{y}} d y

Expand

\frac{y - 1}{e ^{y}} : \frac{y}{e ^{y}} - \frac{1}{e ^{y}}

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

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

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

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

Simplify

= - \frac{y}{e ^{y}}

Add a constant to the solution

= - \frac{y}{e ^{y}} + C

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