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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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