What is the integral of (1-x)(e^{-x}+1) ?

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

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integral of (1-x)(e^{-x}+1)

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

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

Solution steps

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

Expand

(1 - x) (e^{- x} + 1) : e^{- x} + 1 - e^{- x} x - x

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

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

\int 1 d x = x

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

\int x d x = \frac{x ^{2}}{2}

= - e^{- x} + x - (- e^{- x} x - e^{- x}) - \frac{x ^{2}}{2}

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (ln (3 - x))

\frac{d ^{2} y}{d x ^{2}} + \frac{4}{5} \frac{d y}{d x} + \frac{8}{5} y = 0

t \to 1 lim (\frac{5 t - 1}{t + 1})

\frac{\partial}{\partial x} (\frac{a}{s} cos^{2} (x))

n = 1 \sum \infty \frac{n - 6}{n + 6}

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