What is the integral of (3-e^x)e^x ?

The integral of (3-e^x)e^x is -((3-e^x)^2)/2+C

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integral of (3-e^x)e^x

\int (3 - e^{x}) e^{x} d x

- \frac{( 3 - e ^{x} ) ^{2}}{2} + C

Solution steps

\int (3 - e^{x}) e^{x} d x

Apply u-substitution

: \int - u d u

= \int - u d u

Take the constant out:

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

= - \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{u ^{2}}{2}

Substitute back

u = 3 - e^{x}

= - \frac{( 3 - e ^{x} ) ^{2}}{2}

Add a constant to the solution

= - \frac{( 3 - e ^{x} ) ^{2}}{2} + C

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What is the integral of (3-e^x)e^x ?
The integral of (3-e^x)e^x is -((3-e^x)^2)/2+C