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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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