What is the integral of (3x+1)e^{-x+1} ?

The integral of (3x+1)e^{-x+1} is e(-3e^{-x}x-4e^{-x})+C

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integral of (3x+1)e^{-x+1}

\int (3 x + 1) e^{- x + 1} d x

e (- 3 e^{- x} x - 4 e^{- x}) + C

Solution steps

\int (3 x + 1) e^{- x + 1} d x

(3 x + 1) e^{- x + 1} = e^{1} (3 x + 1) e^{- x}

= \int e^{1} (3 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 (3 x + 1) e^{- x} d x

Simplify

= e \cdot \int (3 x + 1) e^{- x} d x

Apply Integration By Parts

: - e^{- x} (3 x + 1) - \int - 3 e^{- x} d x

= e (- e^{- x} (3 x + 1) - \int - 3 e^{- x} d x)

\int - 3 e^{- x} d x = 3 e^{- x}

= e (- e^{- x} (3 x + 1) - 3 e^{- x})

Expand

- e^{- x} (3 x + 1) - 3 e^{- x} : - 3 e^{- x} x - 4 e^{- x}

= e (- 3 e^{- x} x - 4 e^{- x})

Add a constant to the solution

= e (- 3 e^{- x} x - 4 e^{- x}) + C

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What is the integral of (3x+1)e^{-x+1} ?
The integral of (3x+1)e^{-x+1} is e(-3e^{-x}x-4e^{-x})+C