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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

- e^{- x} (1 + x) - e^{- x} : - e^{- x} x - 2 e^{- x}

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

Add a constant to the solution

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

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