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

\int \frac{2 x}{x ^{2} - 3} d x

\frac{1}{2 s ^{2} + 2 s + 1}

arctan (x^{\frac{1}{2}})

\int (x + 3) (x^{2} + 6 x - 7) d x

\frac{\partial}{\partial x} (x^{2} + z^{2})