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

The integral of e^y(y-1) is e^yy-2e^y+C

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integral of e^y(y-1)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

= e^{y} y - 2 e^{y}

Add a constant to the solution

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

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