What is the integral of 1/5 xe^{-x+1} ?

The integral of 1/5 xe^{-x+1} is 1/5 (-e^{-x+1}x-e^{-x+1})+C

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integral of 1/5 xe^{-x+1}

\int \frac{1}{5} x e^{- x + 1} d x

\frac{1}{5} (- e^{- x + 1} x - e^{- x + 1}) + C

Solution steps

\int \frac{1}{5} x e^{- x + 1} d x

Take the constant out:

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

= \frac{1}{5} \cdot \int x e^{- x + 1} d x

Apply Integration By Parts

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

= \frac{1}{5} (- e^{- x + 1} x - \int - e^{- x + 1} d x)

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

= \frac{1}{5} (- e^{- x + 1} x - e^{- x + 1})

Add a constant to the solution

= \frac{1}{5} (- e^{- x + 1} x - e^{- x + 1}) + C

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What is the integral of 1/5 xe^{-x+1} ?
The integral of 1/5 xe^{-x+1} is 1/5 (-e^{-x+1}x-e^{-x+1})+C