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

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

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integral of 2xe^{x-1}

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

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