What is the integral of-6xe^{x+1} ?

The integral of-6xe^{x+1} is -6(e^{1+x}x-e^{1+x})+C

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integral of-6xe^{x+1}

\int - 6 x e^{x + 1} d x

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

Solution steps

\int - 6 x e^{x + 1} d x

Take the constant out:

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

= - 6 \cdot \int x e^{x + 1} d x

Apply Integration By Parts

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

= - 6 (e^{1 + x} x - \int e^{1 + x} d x)

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

= - 6 (e^{1 + x} x - e^{1 + x})

Add a constant to the solution

= - 6 (e^{1 + x} x - e^{1 + x}) + C

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What is the integral of-6xe^{x+1} ?
The integral of-6xe^{x+1} is -6(e^{1+x}x-e^{1+x})+C