What is the integral of+4xe^{-x} ?

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

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integral of+4xe^{-x}

\int + 4 x e^{- x} d x

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

Solution steps

\int 4 x e^{- x} d x

Take the constant out:

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

= 4 \cdot \int x e^{- x} d x

Apply Integration By Parts

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

= 4 (- e^{- x} x - \int - e^{- x} d x)

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

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

Add a constant to the solution

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

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