What is the integral of e^x(x-1) ?

The integral of e^x(x-1) is e^xx-2e^x+C

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integral of e^x(x-1)

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

e^{x} x - 2 e^{x} + C

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

= e^{x} x - 2 e^{x}

Add a constant to the solution

= e^{x} x - 2 e^{x} + C

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