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

The integral of (x+1)/(e^x) is -x/(e^x)-2/(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}} - \frac{2}{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}} - \frac{1}{e ^{x}} - \frac{1}{e ^{x}} : - \frac{x}{e ^{x}} - \frac{2}{e ^{x}}

= - \frac{x}{e ^{x}} - \frac{2}{e ^{x}}

Add a constant to the solution

= - \frac{x}{e ^{x}} - \frac{2}{e ^{x}} + C

\frac{d}{d x} (\frac{x - 1}{x ^{2} - x + 1})

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

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(x^{2} - x y) d x + x d y = 0

\int \frac{x ^{3} + 2}{x ^{2} + 8 x + 7} d x

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