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

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

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

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

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

Solution steps

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

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Simplify

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

Apply Integration By Parts

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

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

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

Expand

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

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

Add a constant to the solution

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

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