What is the integral of 2xe^{1-x} ?

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

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integral of 2xe^{1-x}

\int 2 x e^{1 - x} d x

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

Solution steps

\int 2 x e^{1 - x} d x

Take the constant out:

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

= 2 \cdot \int x e^{1 - x} d x

Apply Integration By Parts

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

= 2 (- e^{1 - x} x - \int - e^{1 - x} d x)

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

= 2 (- e^{1 - x} x - e^{1 - x})

Add a constant to the solution

= 2 (- e^{1 - x} x - e^{1 - x}) + C

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