What is the integral of (9-x)e^x ?

The integral of (9-x)e^x is -e^xx+10e^x+C

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integral of (9-x)e^x

\int (9 - x) e^{x} d x

- e^{x} x + 10 e^{x} + C

Solution steps

\int (9 - x) e^{x} d x

Apply Integration By Parts

: e^{x} (9 - x) - \int - e^{x} d x

= e^{x} (9 - x) - \int - e^{x} d x

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

= e^{x} (9 - x) - (- e^{x})

Simplify

e^{x} (9 - x) - (- e^{x}) : - e^{x} x + 10 e^{x}

= - e^{x} x + 10 e^{x}

Add a constant to the solution

= - e^{x} x + 10 e^{x} + C

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